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Full text of "Principles And Practice Of Qualitative Analysis"

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INTERNATIONAL CHEMICAL SERIES 
Leuis P. HAMMETT, PH.D., Consulting Editor 

+ if 



PRINCIPLES AND PRACTICE 
OF QUALITATIVE ANALYSIS 



The quality of the materials used in 
the manufacture of this book is gov- 
erned by continued postwar shortages. 



A SELECTION OF TITLES FROM THE 
INTERNATIONAL CHEMICAL SERIES 

Louis P. HAMMETT, PH.D., Consulting Editor 



Amsden 

Physical Chemistry for Premedical 
Students 

Arthur 

Lecture Demonstrations in General 
Chemistry 

Arthur and Smith 

Semimicro Qualitative Analysis 

Booth and Damerell 
Quantitative Analysis 

Briscoe 

Structure and Properties of Matter 

Coghill and Sturtevant 
An Introduction to the Preparation and 
Identification of Organic Compounds 



A Laboratory Course in General 
Chemistry 

Dom'eZa 

Mathematical Preparation for Physical 
Chemistry 

Daniels, Mathews, and Williams 
Experimental Physical Chemistry 

Desha 
Organic Chemistry 

Dole- 

Experimental and Theoretical Electro- 
chemistry 

(7*66 

Optical Methods of Chemical Analysis 

Glaastone, Laidler, and Eyrinff 
The Theory of Rate Processes 

Oriffin 

Technical Methods of Analysis 

Hamilton and Simpson 

Calculations of Quantitative Chemical 
Analysis 

Hammett 

Physical Organic Chemistry 
Solutions of Electrolytes 

Henderson and Ferneliua 
Inorganic Preparations 



Problems in Organic Chemistry 

Leighou 

Chemistry of Engineering Materials 

Long and Anderson 
Chemical Calculations 



Millard 
Physical Chemistry for Colleges 

Moore 
History of Chemistry 

Af Orion- 
Laboratory Technique in Organic 

Chemistry 
The Chemistry of Heterocyclio 

Compounds 

Norris ; 

Experimental Organic Chemistry 
The Principles of Organic Chemistry 

Parr 

Analysis of Fuel. Gas, Water, and Lubri- 
cants 

Reedy 

Elementary Qualitative Analysis 
Theoretical Qualitative Analysis 

Rieman, Neuas, and Naiman 
Quantitative Analysis 

Robinson and Oittiland 

The Elements of Fractional Distillation 

Schoch, Felting, and Watt 
General Chemistry 

Snell and Biffen 
Commercial Methods of Analysis 

Steiner 

Introduction to Chemical Thermo- 
dynamics 

Stillwell 

Crystal Chemistry 

Stone, Dunn, and McCullough 
Experiments in General Chemistry 

Thomas 

Colloid Chemistry 

Timm 

General Chemistry 
An Introduction to Chemistry 

Watt 

Laboratory Experiments in General 
Chemistry and Qualitative Analysis 

Williams and Homerberg 
Principles of Metallography 

Woodman 
Food Analysis 



PRINCIPLES AND PRACTICE 
OF QUALITATIVE ANALYSIS 

with Semimicro Laboratory Technique 



PAUL E. SPOERRI, PH.D. 

Associate Professor of Chemistry 
Polytechnic Institute of Brooklyn 

HAROLD WEINBERGER, PH.D. 

Instructor in Chemistry 
College of the City of New York 

ROBERT GINELL, M.S. 

Assistant in Chemistry 
Polytechnic Institute of Brooklyn 



First Edition 

THIRD IMPRESSION 



McGRAW-HILL BOOK COMPANY, INC. 

NEW YORK AND LONDON 
1942 



PRINCIPLES AND PRACTICE OF QUALITATIVE ANALYSIS 

COPYRIGHT, 1942, BY THE 
MCGRAW-HILL BOOK COMPANY, INC. 



PRINTED IN THE UNITED STATES OP AMERICA 

All rights reserved. This book, or 

parts thereof, may not be reproduced 

in any form without permission of 

the publishers. 



COMPOSITION BY THE MAPLE PRESS COMPANY, YORK, PA. 
PRINTED AND BOUND BY COMAC PRESS, INC., BROOKLYN, XT. Y. 



To 
RAYMOND ELLEB KIRK 



PREFACE 

Although the reexamination of the aims of a course is always 
profitable, it is rarely a simple clear-cut problem. This is especially 
true of qualitative analysis, which is one of the oldest courses in the 
chemical curriculum. Although qualitative analysis was originally 
designed as a course to produce industrial analysts, it has, with the 
passage of time, changed its objectives many times. At present we 
teach it to a wide variety of students; some of whom are primarily 
interested in chemistry, some of whom are interested only in related 
fields, and few of whom will ever employ it as a tool. In light of these 
facts, what then should be our aim in teaching qualitative analysis? 

We have given much thought to this problem, and this book is, in 
part, an outgrowth of our ideas. How successful we have been in 
fulfilling our aims is very difficult for us to judge. 

Whereas the first course in chemistry provides the student with a 
background of facts, theories, and laws, we believe that the second-year 
course should have two objectives: (1) to present to the student a 
complete picture of the theory of reactions and -(2) to lead him to 
form an integrated pattern of thought which will effectively guide him 
in his further studies. Hence in writing the. theoretical section of this 
book, our primary aim has been unity of thought. We have tried to 
avoid the presentation of the material in "compartmentalized" form 
even at the risk of disappointing some by omitting their favorite 
topics. We have tried to show how each topic proceeds logically from 
the preceding topics and have attempted to teach by precept the art 
and science of deduction. We believe that qualitative analysis 
should be made interesting and enjoyable, not by tending toward 
oversimplification and popularization but, rather, by appealing to the 
imagination of the student. 

With this aim in view, we have presented the mathematical formu- 
lations, not as something which the student must memorize, but as a 
final result to which the student is led through a clear and logical web 
of thought. On the other hand, we have avoided becoming enmeshed 
in mathematics, since we believe that subjects which can be very well 
understood, for elementary purposes, in a qualitative way, should be 
presented in just such a manner. 



viii PREFACE 

On the experimental side, we believe that a student learns, not 
only by doing, but also by thinking about what he is doing. To further 
this end we have given very few equations throughout the experimental 
section. It is our belief that these should be worked out by the student, 
using the sections on the chemistry of each ion which precedes the 
preliminary experiments. 

Since there is a great tendency on the part of many students to 
utilize any laboratory text as a "cookbook," we have, as far as possible, 
designed the experimental directions so as to minimize their ability 
to do so. Furthermore, we also feel that the methods and apparatus 
used must be simple even though the semimicro technique is employed. 
It is also our conviction that the classical qualitative analysis scheme 
should be taught together with a brief integrated introduction to 
some of the newer techniques. 

Finally, we should always keep in mind this question: What of 
lasting value will the student retain from the course in qualitative 
analysis long after he has forgotten most of the details of the analyses? 

We wish to acknowledge the help received from many of the 
standard and more comprehensive works in the field and numerous 
articles iii the original literature. We also wish to acknowledge the 
encouragement and advice given by Dr. Raymond E. Kirk. Our 
thanks also go to Dr. Gilbert B. L. Smith for helpful discussions and 
for the use of some of his mimeographed material on the solubility 
product; to Dr. H. Mark and Mr. A. L. Davis for assistance in the prep- 
aration of the X-ray photograph; to Paul Becher, Irving Myerson, 
and Mrs. Ruth Weinberger for editorial assistance: to Mrs. Vivian 
Spoerri, Miss Rose Weinberger, Miss Jane Hammerlund, -and Miss 
Kay laconis for the preparation of the manuscript; and to Robert 
Ryder for assistance in the preparation of the section on solutions. 
Finally, credit must be given to the many students who aided immeas- 
urably in working out and testing many of the procedures used. 

PAUL E. SPOERRI, 
HAROLD WEINBERGER, 
ROBERT GINELL. 
BROOKLYN, N.Y., 
October, 1942. 



CONTENTS 

PAGE 

PREFACE vii 

THEORY 

CHAPTER I 
PRINCIPLES OF ANALYSIS 3 

CHAPTER II 
THE PERIODIC AND ANALYTICAL CLASSIFICATION OF THE ELEMENTS 8 

CHAPTER III 
THE ATOM AND THE MOLECULE 24 

CHAPTER IV 
COMPLEX IONS AND MOLECULES 38 

CHAPTER V 
THE KINETIC THEORY OF MATTER 55 

CHAPTER VI 
REACTIONS OF CHEMICAL COMPOUNDS: I, PURE SUBSTANCES 70 

CHAPTER VII 
REACTIONS OF CHEMICAL COMPOUNDS: II, REACTIONS WITH SOLVENTS. ... 92 

CHAPTER VIII 
REACTIONS OF CHEMICAL COMPOUNDS: III, REACTIONS IN SOLUTION .... 121 

CHAPTER IX 

REACTIONS OF CHEMICAL COMPOUNDS: IV, OXIDATION-REDUCTION REAC- 
TIONS 136 

EXPERIMENTAL 

CHAPTER X 
INTRODUCTION 151 

CHAPTER XI . 
MENTAL ATTITUDES 153 

CHAPTER XII 

APPARATUS AND TECHNIQUE 157 

ix 



x CONTENTS 

CHAPTER XIII 

CATION ANALYSIS 167 

Group 1 167 

Group II 172 

Group III 188 

Group IV 203 

Group V 207 

CHAPTER XIV 
INTRODUCTION TO BLOWPIPE ANALYSIS 213 

CHAPTER XV 

ANION ANALYSIS 223 

Group 1 223 

Group II 233 

Group III 244 

Group IV 253 

CHAPTER XVI 
COMPLETE ANALYSIS l 258 

APPENDIX 

UNKNOWNS 263 

LIST OF APPARATUS 265 

SOLUTIONS 267 

INDEX 279 



PART I 

THEORY 



CHAPTER I 
PRINCIPLES OF ANALYSIS 

It is quite natural and commendable that the young student of 
chemistry should be greatly fascinated by the recent achievements of 
creative chemistry the synthesis of vitamins, rubber, dyestuffs, and 
potent medicinals. However, in his eagerness to make his own con- 
tribution, the student is inclined to look upon chemtt-al analysis as an 
unpleasant and unnecessary delay, a drab course maintained in the 
curriculum by academic inertia. Hence it seems appropriate to show 
that such a point of view is not at all justified. In order to build a 
skyscraper, one must first dig deeply into the ground; this seems like 
a retreat from the ultimate goal. Similarly, it cannot be disputed 
that all chemical synthesis rests upon the foundation of chemical 
analysis. 

As an appropriate illustration, let us consider the story of vitamin 
C. This substance, which prevents the previously dreaded disease 
called scurvy, is now inexpensive and is available in any quantity. 
However, the path that led to this end was tortuous. Numerous food- 
stuffs had to be investigated for the presence of this preventive and 
curative principle. Proceeding by elimination, as in any chemical 
analysis, the juices of citrus fruits were found to contain the elusive X. 
In order to isolate the vitamin, the fruit juices had to be taken apart 
and each component carefully examined. After a long process of 
selecting, testing, and discarding impotent materials, crystals were 
finally obtained that had the property of curing scurvy. With this 
achievement, the vitamin had been isolated in pure form, but before 
its final synthesis in the laboratory and its ultimate production in 
a factory, a wide gap, filled with difficulties, had to be bridged. 
Now the problem was to take apart the pure substance. First the 
proportions of carbon, hydrogen, and oxygen were determined. Next 
experiments were performed that differentiated among the various 
possible combinations of these three elements. These experiments 
eventually led to a knowledge of the structural formula of the com- 
pound; the "blueprint" of the molecule. The stage had now been set 
for the synthetic chemist. Chemical analysis had drawn up the indis- 

3 



4 QUALITATIVE ANALYSIS 

pensable blueprint that guided the synthetic chemists' endeavors in 
building up the molecular structure. 

As we have seen, chemical analysis inquires into the composition of 
substances. It asks two questions: what is present and how much of 
each substance. For instance, we might wish to know: Does this 
sample of alluvial sand contain gold? The branch of chemical science 
that enables us to answer this question experimentally and unequivo- 
cally is called qualitative analysis and is the chief concern of this 
volume. However, this indispensable preliminary which answers 
the first question (what) must be complemented by an inquiry into the 
precise quantity of gold present in the sand. The precious metal 
might be definitely present but in such small quantities that its extrac- 
tion would be unprofitable. The science that concerns itself with such 
determinations of quantities is called quantitative analysis. 

As in the case of the proverbial housewife, it may justly be said of 
the analyst that his task never ends. Ever-changing combinations 
of compounds and newly discovered substances and minerals con- 
tinually demand the development of new methods, which often tax the 
ingenuity and skill of the analytical chemist. The astronomer is 
curious about the rare gases in the atmosphere; the criminologist is 
concerned with the composition of a certain cigarette ash; the anxious 
mother wishes to know whether her children's paint s,et contains any 
harmful ingredients; and the department-store buyer would like to 
know whether the silk he contemplates buying is excessively weighted, 
whether the cotton goods in the last shipment contained inferior dyes 
or whether the last lot of drugs conformed to government standards. 

Although all these problems and many more that confront the 
chemist demand a wide variety of methods, it is rather reassuring to 
know that all the methods and principles are based on a single funda- 
mental principle, the principle of subdivision. This is really a chemical 
application of the " divide and conquer" theory so effective in military 
strategy. Let us assume that a clear, aqueous solution contains about 
a dozen cations whose identification is desired. It is known that lead, 
silver, and mercurous ions have one property in common, that of 
forming relatively insoluble chlorides. If, therefore, we add to this 
solution chloride ion in the form of dilute hydrochloric acid, the forma- 
tion of a white precipitate will indicate the presence of at least one of 
these ions. Naturally, such a precipitate could also indicate that a 
mixture of two or of all three of the insoluble chlorides was present. 
In order to obtain further insight into this matter, additional experi- 
ments would be required. This precipitate would be labeled " Group 
I " and reserved for further testing for the presence of the three cations 



PRINCIPLES OF ANALYSIS 5 

possibly present. The solution remaining after the separation of this 
precipitate would now be made slightly acid (by partial neutralization) 
and saturated with hydrogen sulfide gas. This unpleasant but very 
useful operation introduces sulfide ion into the solution. Obviously, 
only those ions having the common property of forming insoluble 
sulfides under these conditions would precipitate. This property is 
shared by a number of ions (bismuth, copper, cadmium, arsenic, etc.), 
and the precipitate, which would be labeled "Group II," must be sub- 
jected to further subdivision procedures. This illustration need not 
be carried further. It is clear that the addition of reagents to the 
unknown in a definite order would yield groups of precipitates that 
contain only certain cations. A common qualitative analysis of the 
cations, recognizes, as a rule, five such groups, as illustrated in the 
following table: 

Group I Silver, lead, mercurous mercury 

Group II Mercuric mercury, lead, bismuth, copper, cadmium, arsenic, 

antimony, tin 

Group III Nickel, cobalt, manganose, zinc, iron, aluminum, chromium 

Group IV Barium, calcium, strontium 

Group V Ammonium, sodium, potassium, magnesium 

The analyst, having separated all the cations into five groups, is, of 
course, still far from the completion of his task. He might be com- 
pared to the postal clerk who has just completed filing all the incoming 
letters according to the states to which they are addressed. This is a 
necessary prerequisite to further division of each pile according to coun- 
ties, towns, and, finally, streets and numbers. Similarly, the mixture 
of precipitates representing a group has to be separated into subgroups 
and finally into compounds containing the individual metals. 

The methods for the final and complete identification of ions should 
also be briefly considered. Lead, which is a constituent of Groups I 
and II, is selected as an example. The presence of this cation is 
indicated whenever the addition of hydrochloric acid produces a white 
precipitate that is soluble in hot water. Scientific thoroughness, 
however, demands further evidence. 

A reference book or the original literature may be consulted, wherein 
will be found a wide variety of experiments concerning the character- 
istic reactions of the lead ion. Obviously, it would not be expedient to 
do all these experiments, and a choice must be made. In this selection, 
we are guided by considering carefully the specificity and sensitivity of a 
test reagent. A reagent would be called specific for lead ion if it gave a 
definite reaction (precipitate, colored solution, etc.) in the presence of 



6 QUALITATIVE ANALYSIS 

lead only but no similar reaction in the absence of lead ion, regardless of 
the presence of any other ion. Such reagents, serving only one master, 
are unknown. Their allegiance is usually divided. Dimethyl 
glyoxime, a reagent for nickel ion, comes closest to being the ideal 
specific reagent. It gives a very characteristic red precipitate with 
nickel in the presence of any or all Group III cations; halogens, how- 
ever, interfere with this test. Therefore, the possibility of interfering 
ions should be carefully considered and reagents chosen that are not 
affected by these possible interferences. Benzidine reagent, for 
- instance, gives a positive color reaction with chromate ion, with 
manganese dioxide, and with lead dioxide. However, the use of benzi- 
dine for the identification of lead ion in Group I is justified, since both 
chromates and manganese dioxide are obviously absent. 

In order to understand the sensitivity of a test, the following simple 
experiments may be considered. Starting with a lead ion stock solu- 
tion of known concentration, a series of more and more dilute solutions 
of known concentration are prepared. One milliliter of each solution 
is placed in a test tube and the tubes arranged in the order of decreasing 
concentration (the first tube, for example, containing 10 mg. of lead 
ion, the second only 5 mg., etc.). An equal quantity of reagent (e.g., 
2 drops), such as potassium chromate, K 2 Cr0 4 , is added to each solu- 
tion. This experiment would show clearly how the concentration of 
the ion would influence the test. In the first tubes, the precipitate 
would be relatively copious, and the identification, therefore, would be 
quite decisive. The volume of precipitate would steadily decrease 
until we reached a test tube where the precipitate would be barely 
visible. This last test would be positive, but a further decrease of the 
lead ion concentration would lead to an uncertain test. The concen- 
tration of this solution would be noted and designated as the limit of 
detection of lead ion with potassium chromate or chromate ion. Such 
concentrations are conventionally expressed in thousandths of a 
milligram, or gamma (7) per drop (0.02 ml.). Repetition of this 
type of experiment with other reagents reveals the fact that the limit 
of detection varies. Let us assume that reagent A has a limit of detec- 
tion of lOy of lead ion, whereas reagent B has a limit of detection of 
0.1 7 of lead ion. Expressed differently, this would mean that B is a 
far more sensitive reagent than A. 

In toxicology, the branch of chemical analysis dealing with poisons, 
reagents must be selected with special reference to their sensitivity. 
Indeed, it would be senseless to test for minute amounts of poisons 
with a reagent of poor sensitivity. This does not mean that we should 
invariably select reagents of the greatest possible sensitivity. Some 



PRINCIPLES OF ANALYSIS 7 

reagents, such as dithizone, are far too sensitive for our laboratory 
work. 

To summarize, in selecting a reagent for a final identification test, 
preference should be given to reagents possessing (1) fair specificity; 
the reagent should not give a similar reaction with any possible con- 
taminants of the test solution and (2) fair sensitivity, permitting 
unequivocal detection. 



CHAPTER II 

THE PERIODIC AND ANALYTICAL CLASSIFICATION OF 
THE ELEMENTS 

The successful practice of analytical chemistry seems at first quite 
arduous, since it appears to demand the memorization of a multitude 
of unrelated and unpredictable facts, such as the solubility of aluminum 
hydroxide in acids and bases or the solubility of arsenous sulfide in 
ammonium polysulfide, etc. Although a certain amount of memoriza- 
tion is indispensable, the task is considerably lightened by a knowledge 
of the periodic arrangement of the elements. This system arranges 
the elements so that it is possible to predict the properties of one ele- 
ment from a knowledge of the properties of another. For instance, 
knowing^that phosphorus has the valences 3 and 5 and that phosphorus 
trioxide and phosphorus pentoxide each form three acids, we can 
predict that arsenic, which is in the same family of elements, will form 
similar compounds that will have analogous properties. 

The periodic classification of the elements was formulated by 
Mendelejeff in 1869. This achievement was the culmination of a 
series of attempts by early nineteenth century chemists to arrange the 
elements on the basis of similarities in their chemical properties. 
Mendelejeff 's original table of the elements has since been modified 
and extended. More recently, the periodic arrangement has been 
shown to depend on a more fundamental quantity than the atomic 
weight, namely, the atotiiic number. 

The Periodic Table. In his original classification, Mendelejeff 
arranged the elements in the order of their increasing atomic weights. 
He found, in so doing, that the properties of the seven elements from 
lithium to fluorine varied systematically in progressing from the first to 
the last named element. The next element, sodium, resembled 
lithium in its properties and reactions, and from this point, the varia- 
tion in properties progressed again, through the same number of 
elements, to chlorine, i.e., each element in this second progression 
resembled a corresponding element in the first series. Upon continuing 
this scheme, Mendelejeff found that all the elements could be arranged 
in a series of horizontal and vertical groups, in which the adjoining 
elements exhibited a gradation of properties. In certain cases, it was 

8 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 



9 



found that the atomic weights of several elements (the factor deter- 
mining their positions) were at variance with the known chemical 
properties. Since the general chemical properties of these elements 
were known with certainty, there were, therefore, two alternatives 
Either the atomic weights, as then known, were incorrect, or the whok 
concept of periodicity was invalid. Boldly, Mendelejeff relied on the 
chemical properties in placing these elements and predicted that the 
atomic weights were incorrect. At other points where no known 
elements could be placed, he left blank spaces and also predicted 
the general properties of the undiscovered elements that would fi1 
into the table properly. Most of these were subsequently discovered 
and isolated. One form of representation of the periodic table is giver 
in Table I. 

TABLE I. THE PERIODIC CLASSIFICATION OF THK ELEMENTS 





Group 




la 


2a 


3a 


4 


5 


6a 


7a 


8 


16 


26 


36 


46 


56 


66 


76 





I 


H 






























He 




1 






























2 


II 


Li 


Be 


















B 


C 


N 


O 


F 


Ne 




3 


4 


















5 


6 


7 


8 


9 


10 


III 


Na 


Mg 


















Al 


Si 


P 


S 


Cl 


A~ 




11 


12 


















13 


14 


15 


16 


17 


18 


IV 


K 


Ca 


Sc 


Ti 


V 


Cr 


Mn 


Fc Co Ni 


Cu 


Zri 


Ga 


Ge 


As 


Sn 


Br 


Kr 




19 


20 


21 


22 


23 


24 


25 


26 27 28 


29 


30 


31 


32 


33 


34 


35 


36 


V 


Rb 


Sr 


Y 


Zr 


Cb 


Mo 


Ma 


Ru Rh Pd 


A K 


Cd 


In 


Sn 


Sb 


Te 


I 


Xe 




37 


38 


39 


40 


41 


42 


43 


44 45 46 


47 


48 


49 


50 


51 


52 


53 


54 


VI 


Cs 


Ba 


La* 


Hf 


Ta 


W 


Ro 


Os Ir It 


Au 


Hg 


Tl 


Pb 


Bi 


Po 





Rn 




55 


56 


57 


72 


73 


74 


75 


76 77 78 


79 


80 


81 


82 


83 


84 


85 


86 


VII 





Ra 


Ac 


Th 


Pa 


U 
























87 


88 


89 


90 


91 


92 























* Elements 57 to 71 are rare-earth elements: 

La Lanthanum 57 Gd Gadolinium 64 

Ce Cerium 58 Tb Terbium 65 

Pr Praseodymium 59 Dy Dysprosium 66 

Nd Neodymium 60 Ho Holmiurn 67 

II Illinium 61 Er Erbium 68 

Sm Samarium 62 Tm Thulium 69 

Eu Europium 63 Yb Ytterbium 70 

Lu Lutecium 71 

Atomic Numbers. Following the publication of Mendelejeff 's 
periodic table, numerous investigations were undertaken to prove or 
disprove his contentions regarding the incorrectness of certain atomic 
weights. In most of these anomalies, the periodic table was shown to 
be correct, and these atomic weights were subsequently revised. 



10 QUALITATIVE ANALYSIS 

TABLE II. ELECTRON DISTRIBUTION IN THE ELEMENTS 





Atomi 




L 


M 


N 





P 


Q 


Ele- 




j 














ment 


nu in 
ber 




















































2i 


2 


3i 


32 


3 


4 


4 


4 


4 


5i 


5 


5 


5 


5 


6 


6 





G 


6 


6 


7i 


H 


1 


1 












































He 


2 


2 












































Li 


3 


2 


1 











































Be 


4 




2 











































B 


5 


JfC 


2 


1 








































C 


6 


shell 


2 


2 








































N 


7 


com- 


2 


3 











































8 


plete 


2 


4 








































F 


9 




2 


5 








































Ne 


10 




2 


6 








































Na 


11 


2 


2 


6 


1 







































Mg 


12 








2 







































Al 


13 


K&ndL 




2 


1 




































Si 


14 


shells 




2 


2 




































P 


15 


complete 




2 


3 




































8 


16 








2 


4 




































Cl 


17 








2 


5 




































A 


18 


' 






2 


6 




































K 


19 


2 


2 


6 


2 


6 





1 
































Ca 


20 















2 
































So 


21 












1 


2 
































Ti 


22 


K andL 








2 


2 
































V 


23 


shells 








3 


2 
































Cr 


24 


complete 








5 


1 
































Mn 


25 












5 


2 
































Fe 


26 












6 


2 
































Co 


27 












7 


2 
































Ni 


28 












8 


2 
































Cu 


29 












10 


1 
































Zn 


30 














2 
































Ga 


31 








M shell 


2 


1 






























Ge 


32 








complete 


2 


2 






























As 


33 














2 


3 






























Se 


34 














2 


4 






























Br 


35 














2 


5 






























Kr 


36 














2 


6 






























Rb 


37 


2 


2 


6 


2 


6 


10 


2 


6 







1 
























Sr 


38 























2 
























Y 


39 


















1 




2 
























Zr 


40 


















2 




2 
























Cb 


41 




X, L, and Af 










4 




1 
























Mo 


42 




sheila 










5 




1 
























Ma 


43 




complete 










6 




1 
























Ru 


44 


















7 




1 
























Rh 


45 


















8 




1 
























Pd 


46 


















10 





























Ag 


47 






















1 
























Cd 


48 






















2 
























In 


49 






















2 


1 






















Sn 


50 






















2 


2 






















Sb 


51 






















> 


3 






















Te 


52 






















> 


4 






















I 


53 






















2 


5 






















Xe 


54 






















2 


6 























THE PERIODIC AND ANALYTICAL CLASSIFICATION 11 
TABLE II. ELECTRON DISTRIBUTION IN THE ELEMENTS. (Continued) 









L 


M 


N 





P 


JQ 




Atomic 
















Ele- 




if 














ment 


num- 


A. 














































ber 




















































2i 


22 


3i 


32 


3s 


4i 


42 


4, 


44 


fr 


62 


5s 


5 4 


56 


61 


62 


6s 


64 


65 


6 


7i 


Cs 


55 


2 


2 


6 


2 


6 


10 


2 


6 


10 




2 


6 









1 














Ba 


56 

































2 














La 


57 


K, L, and M shells complete 












1 






2 














Ce 


58 


2 


2 


6 


2 


6 


10 


2 


6 


10 


1 


2 


6 








2 














Pr 


59 




















2 












2 














Nd 


60 




















3 












2 














11 


61 




















4 












2 














Sm 


62 




K, L.'and M shells 








5 












2 














Eu 


63 




complete 








6 












2 














Gd 


64 




















7 












2 














Tb 


65 




















8 












2 














Dy 


66 




















9 












2 














Ho 


67 




















10 












U 














Er 


68 




















11 












2 














Tm 


69 




















12 












2 














Yb 


70 




















13 












2 














Lu 


71 




















14 






1 






2 














Hf 


72 


2 


2 


6 


2 


6 


10 


2 


6 


10 


14 


2 


6 


2 






2 














Ta 


73 


























3 






2 














W 


74 


K, L, M, and N shells complete 








4 






2 














Re 


75 


























5 






2 














Os 


76 


























6 






2 














Ir 


77 


























7 






2 














Pt 


78 


























9 






1 














Au 


79 


2 


2 


6 


2 


6 


10 


2 


6 


10 


14 


2 


6 


10 






1 














Hg 


80 
































2 














Tl . 


81 
































2 


1 












Pb 


82 


K, L, M, N, and shells complete except 54 and 5 6 




2 


2 












Bi 


83 
































2 


3 












Po 


84 
































2 


4 















85 
































2 


5 












Rn 


86 
































2 


6 















87 


2 


2 


6 


2 


6 


10 


2 


6 


10 


14 


2 


6 


10 






2 


6 










1 


Ra 


88 












































2 


Ac 


89 




































1 








2 


Th 


90 




K, L, M, N, and shells complete except 54 and 5s 






3 








1 


Pa 


91 




































4 








1 


U 


92 




































5 








1 



There were, however, several pairs of elements whose properties 
demanded one arrangement but whose atomic weights, accurately 
rechecked, demanded another. These anomalies proved to be a 
stumbling block for many years and were not cleared up until the 
work of the English physicist Moseley in 1913-1914. Using the 
methods of X-ray spectroscopy, he found that there existed a property 
of the elements more fundamental than atomic weights. This 
property is called the atomic number. It represents the number of 



12 QUALITATIVE ANALYSIS 

positive charges in the nucleus of the atom or, conversely, the total 
number of electrons about the nucleus. In this manner, all the ele- 
ments from hydrogen through uranium may be assigned numbers 
from 1 to 92. 

Electron Shells. It may be seen from the foregoing that the 
chemical behavior of the elements will depend upon their atomic 
number (i.e., number of electrons) and upon some periodic variation 
due to the arrangement of the electrons about the nucleus. The mod- 
ern atomic theory postulates that the nucleus of the atom is surrounded 
by a series of concentric shells containing electrons. It should be 
clearly understood that these are not actual solid shells but are, rather, 
a symbolic representation. The first shell, called the K shell, contains 
two electrons when filled. The second shell, the L shell, contains 
eight electrons when completed, etc. The number of electrons in the 
outermost shell determines the chemical properties of the element. 
Table II shows the electron distribution in shells for each element. 

Vertical Relationships. The construction of the periodic table 
(Table I) is such that the elements in the vertical groups or families 
bear a strong chemical resemblance to each other. For example, the 
members of Group la (lithium, Li, sodium, Na, potassium, K, rubid- 
ium, Rb, and cesium, Cs) exhibit a usual valence of positive 1. Their 
hydroxides, which have the general formula MOH, arc all strong bases 
and are quite soluble in water. The strongest base is cesium hydroxide, 
and the weakest is lithium hydroxide. The general formula of the 
hydroxides of the elements of Group 2a (calcium, Ca, barium, Ba, and 
strontium, Sr) is M(OH) 2 . These are also strong bases but have only a 
limited solubility in water. 

The base-forming character of the elements decreases through the 
next several groups through the amphoteric, or transitional, elements 
to the acid-forming elements. In Group 76, for example, are the 
elements fluorine, chlorine, bromine, and iodine, the so-called halogens. 
The compounds of the halogens with hydrogen are fairly strong acids in 
aqueous solution, decreasing in stability to HI. These elements, with 
the exception of fluorine, also form oxygen acids, which are active 
oxidizing agents. 

The remaining vertical groups exhibit similarities of the same type. 
The elements of Group 66 sulphur, S, selenium, Se, and tellurium, 
Te all form similar hydrogen compounds having the general formula 
H 2 X. These are all gases with foul odors and are slightly soluble in 
water with the formation of weak acids. They function as precipitants 
with many metallic ions and are also good reducing agents. They 
form a series of strong oxygen acids and give rise to oxides of formula 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 13 

X02 and X0 3 . The last member of the group, polonium, Po, is 
one of the radioactive elements and is quite rare. Its reactions have 
not been fully investigated, but it probably reacts in the same fashion 
as the other members of the group. Similarly, the reactions of phos- 
phorus, arsenic, antimony, and bismuth follow the same general 
family relationships. It should be remembered that there is a grada- 
tion of properties from the first element in a vertical group to the last, 
and therefore the first may bear only a superficial relationship to the 
last element. However, the elements in the middle of a group will 
resemble each other more closely. 

One more important point must be considered in this section. 
Hydrogen, in Group 1, appears to fall in two families. First, in some 
of its reactions, it resembles the alkali metals of Group la and forms 
stable compounds with the elements of Group 76, having the 
formula HX, and with the elements of Group 66, where the formula 
is H 2 X. In these reactions, hydrogen has a valence number of positive 
1. Similarly, it also appears to belong to Group 76, since it forms 
hydrides with the alkali metals of formula MH and with Group 2 
metals, where the formula is MH 2 . Here hydrogen has a valence 
number of negative 1. 

In general, the first member of each vertical group exhibits some 
anomalous reactions. This is generally attributed to the smaller 
atomic volume of the clement. The boiling point of the hydrogen 
halides of Oroup 76, varies thus : 

HI = -36C., HBr = -68C., HC1 = -83C., HF = +19.4O. 

It can be seen that the boiling point of hydrogen fluoride is anomalous. 
This is attributed to the small volume and extreme reactivity of fluorine. 
HF reacts with another HF to form the dimer H 2 F 2 . Other examples 
of the anomalous behavior of the first members are oxygen in Group 66 
and nitrogen in Group 56, both of which are gases, whereas the other 
members of both groups are solids. 

Horizontal Relationships. In addition to the strong chemical 
resemblances noted among the members of the vertical groups or 
families of elements, there exist, also, marked similarities among a 
number of elements in the various horizontal series in the periodic 
table. In the first three series, those beginning with hydrogen, 
lithium, and sodium, respectively, the adjoining elements exhibit little 
similarity in properties but demonstrate, rather, a marked difference 
in progressing from one to the other. However, in the fourth series, 
that which begins with potassium, the first few elements, notably 
potassium and calcium, exhibit the same marked difference, but this 



14 QUALITATIVE ANALYSIS 

gradation in properties ends with a series of elements that show many 
similarities in properties. This series continues to Group 8 (iron, 
cobalt, and nickel), where the similarities among the members of this 
group are so pronounced that they are placed together in one group and 
considered as a single element in the periodic arrangement. These 
elements are commonly known as transitional elements. From this 
point, the elements return to the point where marked gradations in 
properties*are again noted. 

In the first two (short) series, the elements gradually decrease in 
metallic character in progressing from Group la across to Group 76. 
However, the " trough" elements in the first long series are all metals. 
The same may be said regarding the second long series. In the third 
long series (Series VI), the trough character is greatly extended, and 
from element 57 (lanthanum) to element 72 (hafnium), there are 14 
elements whose properties are remarkably similar (the rare-earth 
metals). In fact, they are so closely related that they are usually 
found associated in their ores and complex minerals and are exceedingly 
difficult to separate. They have all been isolated, in the form of their 
salts, by laborious and elaborate fractional-crystallization procedures. 

This same close relationship is also shown among the elements of 
Group 8, where nickel is more frequently found associated with 
cobalt in ores than with palladium and where platinum is more fre- 
quently found with osmium and iridium than with palladium (see 
Table I). The student should note carefully that, although the verti- 
cal relationships of the elements are important, in many cases, par- 
ticularly among the trough elements of the long series, the horizontal 
relationships are just as important and sometimes are even more 
important. 

Diagonal Relationships. A third type of relationship among the 
elements, in addition to the two types discussed above, is worthy of 
mention. This is the diagonal relationship that the elements bear to 
each other in the periodic classification. It appears that certain 
elements usually have some chemical resemblance to the elements 
diagonally below them to the right. This is especially true of the 
elements of the first series. Beryllium, although it has a valence num- 
ber of positive 2 and forms a chloride BeCl 2 , strongly resembles 
aluminum, which has a valence of positive 3 and forms a chloride 
Aids, rather than magnesium, which is just below it in Group 2a. 
In fact, beryllium is rather difficult to separate from aluminum. 
Another example is that of lithium, which resembles magnesium in its 
reactions. Similarly, boron is a nonmetal like silicon, its diagonal 
partner, rather than a metal like aluminum, which is in the same 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 



15 



family. Other resemblances of the same sort are to be found through- 
out the periodic table, although these are not so pronounced as the 
examples cited. 

ANALYTICAL GROUPING OF THE ELEMENTS 

Twenty-one elements are generally included in the conventional 
schemes of qualitative analysis. In addition, two different valence 
forms of mercury (mercurous and mercuric ions) are considered indi- 
vidually, and lead, by virtue of the slight solubility of its chloride, is 
considered in two separate groups. The ammonium ion, which 
resembles the alkali metals closely in its chemical properties, is also 
included, making a total of 25 cations (Table III). For analytical 

TABLE III. ELEMENTS INCLUDED IN THE USUAL SCHEMES OF CATION ANALYSIS 
H He 



Li Be 



B C N O F Ne 



Na I 


tfg 






Al 


Si P 


S Cl A 
















K 


Ca 


Sc Ti V Cr MnFe Co Ni Cu 


Zn 


Ga 


Ge As 


Se Br Kr 


Rb 


Sr 


Y Zr Cb Mo Ma Ru Rh Pd Ag 


Cd 


In 


Sn Sb 


Te I Xe 


Cs 


Ba 


La Hf Ta W Re Os Ir Ft Au 


Hg 


Tl 


Pb Bi 


Po Rn 



Ra Ac Th Pa U 



purposes, these metals are divided into five groups, based upon their 
reactions with various reagents. This method is based primarily 
upon the solubilities of various compounds of the ions and is, therefore, 
an arbitrary arrangement. Obviously, other methods of grouping 
could, and have been, worked out, but the arrangement given in this 
text is, with some variations, the most convenient and that which is 
almost universally used. It should be noted, too, that this arrange- 
ment is not based entirely upon the periodic system, but a study of the 
various figures will show that a close similarity does exist among the 
vertical, horizontal, and diagonal relationships in the periodic table 
and the analytical grouping of the metals. 



16 QUALITATIVE ANALYSIS 

The Groups. The elements are divided as follows: 

Group I contains those metals which form insoluble or sparingly 
soluble chlorides. These are silver, lead, and mercurous mercury. 
The group reagent is Cl~~, usually in the form of dilute hydrochloric 
acid. Since lead chloride is slightly soluble in water, sufficient lead 
ion remains in solution to make it necessary to test for the cation again 
in Group II. Table IV shows the elements that form insoluble chlo- 
rides and their relationships in the periodic table. 

Group II consists of those metallic ions which yield sulfides that are 
insoluble in dilute acids. These are usually precipitated, after the 
removal of Group I, by the use of hydrogen sulfide gas in solutions 
about 0.25 to 0.5JV with respect to hydrochloric acid. The elements 
thus precipitated are: mercuric mercury, Hg, lead, Pb, bismuth, Bi, 
copper, Cu, cadmium, Cd, arsenic, As, antimony, Sb, and tin, Sn. 
The group is further subdivided into subgroups A and B by the solvent 
action of ammonium polysulfide on the sulfides of the three last 
elements (Tables V, VI, VII). 

Group III comprises those cations that yield sulfides that are soluble 
in dilute acids but that are insoluble in neutral or alkaline medium 
and those ions that form insoluble hydroxides with ammonia. This 
group, therefore, is precipitated by hydrogen sulfide from ammo- 
niacal solution (ammonium sulfide). The relation of these elements in 
the periodic table is illustrated in Tables VIII, IX, X. 

Group IV metals are those which form soluble sulfides under the 
conditions given above but which yield other insoluble salts after the 
separation of the previous groups. The alkaline earth elements 
(calcium, barium, and strontium) comprise this group, as shown in 
Table XI. 

Group V contains those elements not precipitated by any of the 
reagents previously used. These are magnesium, the alkali metals, 
sodium and potassium, and ammonium ion. These are generally 
identified individually in the final filtrate, although the test for 
ammonium ion is usually performed at the beginning of an analysis 
(Table XII). 

Use of the Analytical Tables. In attempting to predict the chem- 
ical properties of any element, it should always be borne in mind that 
there are three types of similarities (vertical, horizontal, and diagonal 
relationships) which must be weighed and balanced before a decision is 
made. In Tables IV to XII, it has been shown how the elements 
react to certain specific and commonly used reagents. For example, if 
a chemist should have occasion to analyze a sample suspected of 
containing germanium, what could he say about the likely properties 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 17 

TABLE IV. GROUP I METALS 

ELEMENTS THAT FORM INSOLUBLE OR SPARINGLY SOLUBLE CHLORIDES 
H He 



Li Be 



B C N F Ne 



Na Mg 



Al Si P S Cl A 



K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 



Rb Sr Y Zr Cb Mo Ma Ru Rh Pd Ag Cd In Sn Sb Te I Xe 



Cs Ba La Hf Ta W Re Os Ir Pt Au Hg ml Pb Bi Po Rn 



Ra Ac Th Pa U 

Light block indicates element that is not considered in the usual schemes of cation 

analysis. 

TABLE V. GROUP II METALS 

ELEMENTS THAT FORM INSOLUBLE SULFIDES IN 0.25 TO 0.5# ACID SOLUTIONS 
H He 



Li Be 



B C N F Ne 



Na Mg 



Al Si PS Cl A 



Rb Sr Y Zr Cb 



Cs Ba La Hf Ta 



Cr MnFe Co Ni 


Cu 


Zn 


Ga 
In 




|Ge) As 
Sn Sb 
Pb Bi 




Se 
Te 


Mo Ma Ru Rh Pd 
W Re Os Ir Pt 





Cd 
Hg 


Po 


Au 



Br Kr 



I Xe 



Po Rn 



Ra Ac Th Pa U 

Light blocks indicate elements that are not considered in the usual schemes of 
cation analysis. Circles indicate elements previously removed. 



QUALITATIVE ANALYSIS 

TABLE VI. GROUP IIA ELEMENTS 

ELEMENTS THAT YIELD SULFIDES INSOLUBLE IN AMMONIUM POLYSULFIDE 
H He 



Li Be 



B C N F Ne 



Na Mg 



Al Si P S Cl A 



K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 



Rb Sr Y Zr Cb Mo Ma 



Cs Ba La Hf Ta W 



Ru Rh Pd (Ag) Cd 



Ir Pt Au 



Hg 



In Sn Sb Te I Xe 



Tl) Pb Bi Po Rn 



Ra Ac Th Pa U 

Light blocks indicate elements that are not considered in the usual schemes of 
cation analysis. Circles indicate elements previously removed. 

TABLE VII. GROUP IIB ELEMENTS 
ELEMENTS THAT YIELD SULFIDES SOLUBLE IN AMMONIUM POLYSULFIDE 

H He 



Li Be 



B C N F Ne 



Na Mg 



Al Si P S Cl A 



K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 



Rb Sr Y Zr Cb 



Cs Ba La Hf Ta 



Mo Ma 



Ru Rh Pd Ag Cd In Sn Sb Te I Xe 



W 



Re Os Ir Pt Au| Hg Tl Pb Bi Po Rn 



Ra Ac Th Pa U 

Light blocks indicate elements that are not considered in the usual schemes of 

cation analysis. 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 19 

TABLE VIII. GROUP III ELEMENTS 
ELEMENTS THAT YIELD INSOLUBLE SULFIDES OR HYDROXIDES WITH AMMONIUM 

SULFIDE AND AMMONIUM HYDROXIDE 
H He 



Li Be 



B C N O F Ne 



Na Mg 
K Ca 
Rb Sr 
Cs Ba 
Ra 




Al 


Si P S Cl A 
Ge As Se Br Kr 
Sn Sb Te I Xe 

@1 Po ~" Rn 


Sc Ti V 
Y Zr Cb 
La* Hf Ta 
Ac Th Pa 


Cr Mn Fe 


Co Ni @ 


Zn 




Ga 
In 




M< 


) Ma Ru 

_T 

Re (5s) 


] 


@ 

U 


Ir Pt Au 







Light blocks indicate elements that are not considered in the usual schemes of 
cation analysis. Circles indicate elements previously removed. 

* Including rare-earth elements. 

TABLE IX. GROUP IIIA P^LEMENTS 
ELEMENTS THAT FORM INSOLUBLE HYDROXIDES WITH AMMONIUM SULFIDE AND 

AMMONIUM HYDROXIDES 
H He 



Li Be 



B C N O F Ne 



Na Mg 



Al Si P S Cl A 



K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 



Rb Sr 



Cs Ba 



Y Zr 



La* Hf 



Ra Ac 



Th 



Cb Mo Ma Ru Rh Pd Ag Cd In Sn Sb Te I Xe 



Ta W Re Os Ir Pt Au Hg TI Pb Bi Po Rn 



Pa U 



Light blocks indicate elements that are not considered in the usual schemes of 
cation analysis. Elements previously removed are not included in this figure. 

* Including rare earths. 



20 



QUALITATIVE ANALYSIS 



TABLE X. GROUP lllB ELEMENTS 
ELEMENTS THAT FORM INSOLUBLE SULFIDES WITH AMMONIUM SULFIDE AND 

AMMONIUM HYDROXIDE 
H He 



Li Be 



B C N O F Ne 



Na Mg 



Al Si P S Cl A 



K Ca Sc Ti V Cr |Mn Fe Co Ni] Cu |Zn JGa Ge As Se Br Kr 



Rb Sr Y Zr Cb Mo Ma Ru Rh Pd Ag Cd In Sn Sb Te I Xe 



Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po Rn 



Ra Ac Th Pa U 



Light blocks indicate elements that are not considered in the usual schemes of 
cation analysis. Elements previously removed arc not included in this figure. 

TABLE XI. GROUP IV ELEMENTS 
ELEMENTS PRECIPITATED BY AMMONIUM CARBONATE AND AMMONIUM CHLORIDE 

(After the separation of the previous groups; other methods also used) 
H He 



Li Be 



B C N F Ne 



Na Mg 



Al Si P S Cl A 



K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 



Rb 



Cs 



Sr 



Ba 



Ra 



Y Zr Cb Mo Ma Ru Rh Pd Ag Cd In Sn Sb Te I Xe 



La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po Rn 



Ac Th Pa U 



Light block indicates elements that are not considered in the usual schemes of 

cation analysis. 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 21 

TABLE XII. GROUP V ELEMENTS 
ELEMENTS THAT FORM SOLUBLE COMPOUNDS WITH THE REAGENTS USED 

PREVIOUSLY 
H He 



Be 



Rb 



Cs 



Na Mg 



K 



B C N O F Ne 



Al Si P S Cl A 



Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 



Sr Y Zr Cb Mo Ma Ru Rh Pd Ag Cd In Sn Sb Tc I Xo 



Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po Rn 



Ra Ac Th Pa U 

Light blocks indicate elements that are not considered in the usual cation-analysis 

schemes. 

TABLE XIII. ELEMENTS NOT INCLUDED IN THE PREVIOUS TABLES 



Li Be 

Na Mg 



|B C N O F 



Al 



Si P S Cl 



K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se 



Rb Sr Y Zr Cb Mo Ma Ru Rh Pd Ag Cd In Sn Sb Te 



Br 



Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi jPoj Rn 



He 



Ne 



Kr 



Xe 



Ra Ac Th Pa U 

Light block indicates rare gases chemically inert (as usually understood). 

Heavy block indicates elements usually encountered as anions. 

Dotted block little is known of chemistry, although probably similar to Se 

and Te. 

Circle very active chemically special analytical methods employed. Always 
present in aqueous systems. 



22 , QUALITATIVE ANALYSIS 

of this element and its compounds before consulting a reference book? 
Looking at Table I, he would find that it is in Group 46, situated 
between silicon and tin. Silicon is a nonmetal, and tin is a metal; 
germanium should then be somewhere in between. In the horizontal 
series in which it occurs, it lies between zinc, an amphoteric metal, and 
arsenic, which is less metallic and also amphoteric; therefore he could 
safely predict that germanium will be amphoteric. From Table VI, 
we see that it falls into the analytical Group I IB. We could also, from 
a knowledge of the chemistry of its surrounding elements, say that it 
should hydrolyze in water forming an oxy compound. We have now 
deduced some pertinent information about this element, and we can 
proceed with the analysis of the other groups but expect to have diffi- 
culties in group 115. With the knowledge at hand, a reference book 
should now be consulted for the exact modifications to be introduced 
in the group 115 analysis. 

READING REFERENCES 

HILDEBRAND, J. H.: "Principles of Chemistry," 4th ed., Chap. XVTT, The Mac- 

millan Company, New York, 1940. 
MELDRUM, W. B., and F. T. GUCKER: "Introduction to Theoretical Chemistry," 

Chap. I, American Book Company, New York, 1936. 
MOORE, F. J.: "A History of Chemistry," 2d cd., Chap. XVI, McGraw-Hill 

Book Company, Inc., New York, 1931. 
QUAM, M. B., and G. N. QUAM: Types of Graphic Classification of the Elements, 

J. Chem. Educ., Vol. 11, January, April, and May, 1934. 
WEEKS, M. E.: "The Discovery of the Elements," Chap. XIV and XV, Mack 

Printing Company, Easton, Pa., 1935. 

QUESTIONS 

1. Using a periodic chart of the elements, predict the formulas of the following 
compounds of the element whose atomic number is 38: Oxide, chloride, sulfate, 
nitrite. 

2. Using a periodic chart of the elements, predict the formula of the following 
compounds of the clement whose atomic number is 81: Oxides, chloride, nitrate, 
phosphate. 

3. With what element or elements would you expect cadmium to be asso- 
ciated in ores? 

4. How many electrons are there in the outermost shell of the following 
elements: vanadium, rhenium, beryllium, aluminum, radium. 

6. Given that the density of sulfur is 2.07 and that the density of tellurium 
is 6.24, what would you predict that the density of selenium would be? Compare 
this value with that given in the literature, and calculate your percentage error. 

6. Given that the melting point of potassium is 62.3C. and that the melting 
point of cesium is 28.5C., what would you predict that the melting point of rubid- 
ium would be? Compare this value with that given in the literature, and calculate 
your percentage error. 



THE PERIODIC AND ANALYTICAL CLASSIFICATION 23 

7. If you suspected that you had columbium in an unknown, in what ana- 
lytical cation group would you look for it? 

8. In what analytical group would gold precipitate if it were present in one 
of your unknowns? 

9. An analyst who specializes in the analysis of the noble metals is concerned 
Chiefly with what analytical groups? 

10. What arrangement of the electrons characterizes a rare gas? 



CHAPTER III 
THE ATOM AND THE MOLECULE 

Although no one, to date, has seen an individual atom, scientists 
have nevertheless been able to gather a multitude of data and informa- 
tion about them and, as a consequence, have tried to imagine what 
atoms look like. In the earlier pictures, the atom consisted of a small 
positively charged nucleus, around which rotated negatively charged 
electrons in some sort of orbits. 

However, this picture was not quite satisfactory, since it was found 
that an atom constructed on this plan would not behave as a real atom 
does. For instance, a real atomic system emits light of a definite 
wave length (energy) when heated. In the simple model mentioned, 
the rotating electron would continuously emit energy of changing 
wave length and would revolve in a spiral that became smaller and 
smaller until the electron struck the nucleus. 

The frequency (the number of waves per second) is related to the wave length 
(the length of each wave) by velocity, since number (of waves) times centimeters 
gives centimeters per second, which has the dimensions of velocity, This can be 
represented by the formula 

v\ = v 

where v = velocity 
v = frequency 
X = wave length 

The Hydrogen Atom. Niels Bohr formulated an empirical solution 
to this problem. His model is similar to this simple one except that 
the electrons are permitted to travel only in orbits of certain fixed 
diameters and not in all possible orbits (and therefore can never fall 
into the nucleus). He arrived at this concept quite empirically, i.e., 
he found that if he imposed this condition, called the quantum condi- 
tion, on a model built on classical lines, the model would behave like a 
real atom. In the picture that he proposed, an electron revolving 
about the nucleus would not emit energy (as light) but on jumping 
from one orbit to another orbit, would emit light. The frequency of 
this light could be calculated simply. The energy that the electron 
possessed at each level could be calculated from classical mechanics, 
and the difference between the energies at these two levels would 

24 



THE ATOM AND THE MOLECULE 



25 



represent the energy emitted as light of a definite frequency. The 
relationship between the energy and frequency is simple. The energy 
equals the frequency times h (Planck's constant) or 

E = hv 
where E = energy 

v = frequency 

h = Planck's constant 

It was now possible to draw a schematic picture of the hydrogen 
atom according to this theory. In going from a higher energy level to 
a lower energy level, the electron emits light or energy; in going from a 
lower energy level to a higher one, it absorbs energy or light. When a 
vapor is heated, the electrons jump to higher levels and absorb energy, 
and in falling back to lower levels, they emit light, which, when 

2nd Orbft^ tfZrJ. Orbit 




1st Orbit* ^Nucleus 
FIG. 1. Schematic pic- 
ture of some possible circu- 
lar orbits (energy levels) of 
the electron in the hydrogen 
atom. 




FIG. 2. Schematic picture of 
some circular and elliptical orbits of 
the electron in the hydrogen atom. 



analyzed spectroscopically, appears as a series of lines (of fixed posi- 
tion for each element) of characteristic wave length. 

Later it was realized that when one body revolves around another 
body, it does not usually move in a circular path but, rather, in an 
elliptical one. This is known from a study of both theoretical mechan- 
ics and practical astronomy. It has long been known that the orbits 
of the planets around the sun are elliptical rather than circular, with 
the sun at one of the focuses. This refinement of the Bohr theory 
resulted in a change in the concept of the atom. The orbits were now 
elliptical, as shown in Fig. 2. 

This newer concept explained fairly adequately the structure of the 
hydrogen atom, but it was not so successful when more complicated 
atoms were considered. Furthermore, it was not clearly understood 
why this quantum condition was necessary. 



26 



QUALITATIVE ANALYSIS 



An explanation of this problem was soon forthcoming. De Broglie 
and Schroedinger changed the concept of the atom by inventing a new 
tool, quantum mechanics, and endowed the empirical relationship of 
Bohr with a theoretical explanation. By the use of higher mathe- 





Fio. 3. Circular standing wave: 
permissible orbit. 



FIG. 4. Circular wave interferes 
with itself: iionpermissible orbit. 



Aphelion- 



-Lowee-f- 
velocity 



\ 



natics (beyond the scope of this book), they derived a basis for Bohr's 
quantum condition. On a simplified basis, the reason for this condition 
;an be made intelligible. Let us first consider that the electron, in 
addition to being a particle, is also a wave trav- 
eling in a circular orbit around a nucleus, as in 
Fig. 3. This wave is continuous, a sort of a 
stationary wave. If, however, the circumfer- 
ence of the orbit is slightly decreased, stationary 
waves cannot be produced, but each wave will 
interfere with the preceding wave (as in Fig. 4) 
and produce beats and interference. Therefore, 
it is evident that stationary waves of a certain 
wave length will exist only in certain orbits of 
fixed radii. This is a very simplified two-di- 
mensional picture of Schroedinger's explanation 
of Bohr's quantum condition. De Broglie's 
and Schroedinger's theories, therefore, changed 
the hitherto material particle, the electron, 
into an electrical-charge density or a wave in 
an electrical cloud. This latter statement may 
eem almost unintelligible, but it has meaning. 

M. Born has given a new interpretation to this concept. Accord- 
ig to him, "the electrical charge density in an electrical cloud," con- 
idered statistically is nothing more than the probability of finding a 




-Nucleus 



velocity 

FIG. 5. Path of 
lectron traveling in 
n elliptical orbit 
round the nucleus. 



ATOM AND THE MOLECULE 



27 



material particle, an electron, in any position around a nucleus. One 
way of interpreting this statement is as follows. Let us look at an 
electron traveling in an elliptical orbit around a nucleus. The speed of 
the electron will not be constant because, according to Kepler's laws of 
motion, the area swept by a body moving around another one is the 
same for each period of time. Therefore, the electron will move most 
rapidly when it is nearest the nucleus (at the perihelion) and least 
rapidly when it is farthest from the nucleus (at the aphelion). Accord- 
ing to Einstein's theory of relativity, the more rapidly a body moves 
the greater its mass. This effect, although only very slight at ordinary 
velocities, is enhanced as the body approaches the velocity of light. 





FIG. 6. Precession of ,tho 
elliptical orbit of the electron. 



Fia. 7. Limits of a processing 
orbit. 



Therefore, since the electron moves with a very high velocity, any 
change in this velocity will give rise to a pronounced effect on the mass. 
The electron-, therefore, has a greater mass because of its greater 
velocity when it is nearer the nucleus. The curvature of the orbit 
depends upon the mass and velocity of the electron, and, as a conse- 
quence, the path the electron returns upon will not be the mirror 
image of the path it originally traversed, and the elliptical orbit will 
precess (see Fig. 6). 

The electron will therefore never approach closer to the nucleus 
than circle a (Fig. 7) or go farther than circle 6, and we may consider 
the space between a and b as being an electrical cloud of variable 
electrical density, or we may consider the statistical probability of 
finding the electron in any small area between a and 6. As is evident 
from the drawing, the probability of finding an electron in a small area 
near the nucleus is greater than finding it far away from the nucleus. 

Another method of describing the atom is by means of Heisen- 
berg's matrix mechanics. Unfortunately, no one has yet been able to 



28 



QUALITATIVE ANALYSIS 



derive a model or picture of the atom from his mathematical equations, 
which are quite complicated and which will, as a consequence, not be 
considered. 

We see, therefore, that an electron is a many-sifted object that 
assumes varied aspects, depending upon one's vantage point. The 
objective, then, determines which aspect of the electron is used. 

The structure of the nucleus, the other half of the hydrogen atom, 
(if it has a structure?) is a complete mystery to science. All that can 
be said about the hydrogen nucleus, or proton, is that it is singly posi- 
tive in charge, has a mass of about 1.65 X 10~ 24 gram and a radius of 
the order of 10~ 13 cm. 

Thus far, the physical aspects of the structure of the hydrogen atom 
have been considered. Since our subject of study is elementary 
chemistry, we shall consider henceforth the atom in two relatively 
simple ways, which will serve to explain most of the phenomena that 
we shall encounter. In the first picture, the nucleus is represented by a 
dot and the electrons by other dots around it, as in Fig. 8. 



o e 





Hydrogen Helium 

Lithium Fluorine 

FIG. 8. Symbolic representation of electronic structure. 

In the second picture, the nucleus and all the electrons in the inner 
shells are represented by the symbol of the element, and the electrons 
in the outermost shell are represented by dots. If any electrons were 
originally present and were later removed because of reactions, no dots 
are indicated (see lithium atom and lithium ion). The magnitude of 
the charge on the atom is indicated by the appropriate symbols (+ 
or-). 



H- 

Hydrogen 
atom 

-.Cl- 

Chlorine 
atom 



H+ 

Hydrogen 
ion 



Chloride 
ion 



He- 

Helium 
atom 



Li- 

Lithium 
atom 



Lithium 
ion 



Na- 

Sodium 
atom 



Na+ 

Sodium 
ion 



Cu: Cu++ 



Copper 
ato'm 



Cuprio 
ion 



The student should always remember that these are only symbols 
and not actual pictures of atoms. 

Other Atoms. As was pointed out in Chap. II, the elements 
exhibit a periodic gradation in properties. This can be explained on 



THE ATOM AND THE MOLECULE 



29 



an electronic basis. Hydrogen, with an atomic number of 1, has one 
electron revolving around the nucleus, and it may be represented as in 




Hydrogen atom 

Atomic number ~i 

FIG. 9. 




Helium atom 

Atomic number^-2 

FIG. 10. 




Lithium atom 
Atomic 

FIG. 11. 



Fig. 9. Helium, with an atomic number of 2, has two electrons 

revolving around it, as in Fig. 10. The next element is lithium, which 

has an atomic number of 3. It could be written as in Fig. 

12, but it is known that lithium is somewhat similar to 

hydrogen in its reactions, and Fig. 9 does not resemble 

Fig. 12 in the slightest. If, however, we now write lithium 

as in Fig. 11, it would bear some resemblance to Fig. 9; 

both representations have a lone electron in their outer 

ring. The following elements to neon, atomic number 10, would then 

appear as in Fig. 13. 




Fio. 12. 





Beryllium 
Atomic No.4- 



Boron 
Atomic No. 5 





Carbon 
Atomic No. 



Nitrogen 
Atomic No. 7 





Oxygen Fluorine Neon 

Atomic No. 8 Atomic No. 9 Atomic No. 10 

FIG. 13. Symbolic representation of the remaining members of the second series. 

Neon, atomic number 10, is an inert gas and resembles helium; the 
next element sodium, atomic number 11, resembles lithium in its 
reactions and should evidently have a lone electron in its outer ring 
and could be represented as in Fig. 14. The same idea may be applied 
in the formulation of the pictures of all the other elements. However, 
since this symbolism becomes a little cumbersome with the elements of 



30 



QUALITATIVE ANALYSIS 



higher atomic number and since only the electrons in the outer ring 
usually participate in chemical reaction, the briefer symbols shown 
previously are more commonly used. 

The Hydrogen Molecule. Up to this 
point, we have treated only isolated atoms. 
Let us consider what would occur if two atoms 
of hydrogen approached one another. At 
relatively great distances, nothing would 
happen, but as the distance between the 
atoms grew progressively smaller, a point 
would be approached when the two nuclei 
would be so close that each would influence 
the motion of the electron surrounding the 
other nucleus. The aberrations in the motion of the electrons will 
become greater and greater, and the electrons will soon be driven into 
new orbits, in which they circle both nuclei (see Fig. 15). This pic- 




Soclfum 
Atomic No. 11 
FIG. 14. 






FIG. 15. Hydrogen mole- 
cule. 



FIG. 16. Hydrogen mole- 
cule: Schroedingcr picture. 



ture may also be considered from the Schroedingcr point of view, where 
the orbits now become electrical clouds (Fig. 16). In following the 
briefer symbolism used previously, the hydrogen molecule would 
appear as 

H:H 

Hydrogen molecule 

In this case, the electrons belong to both nuclei; as can be seen in Fig. 
17, the hydrogen molecule resembles the helium atom in that both have 





Hydrogen Molecule 



Helium Atom 



FIG. 17. 



two electrons in their outer ring. The arrangement of the electrons 
about the nucleus in rare gases is peculiarly stable, and all atoms tend 
to arrange themselves so that they may assume arrangements as similar 



THE ATOM AND THE MOLECULE 31 

to these elements as possible. Hydrogen at room temperature is 
completely in the form of diatomic molecules, whereas helium, neon, 
and the other rare gases exist in the monatomic state. 

The Homopolar Bond. The type of bond that exists between two 
atoms of hydrogen is called the homopolar or nonpolar bond. This 
type of bond may also exist between dissimilar atoms, as in hydrogen 
chloride, carbon tetrachloride, methane, etc. 

:C1: H 

H:C1: : C1:C:C1: H:C:H 

":C1:" ii 

Hydrogen Carbon Methane 

chloride tetrachloride 

This type of bond is often represented by a single line. 

Cl H 

H Cl Cl C Cl H C H 



Cl 



l H 

Hydrogen Carbon Methane 

chloride tetrachloride 

The Heteropolar Bond. A second type of bond is called the hetero- 
polar bond. If two atoms approached each other, as previously 
described but in this case had different attractive power for electrons, 
the atom with the greater attraction for an electron would remove it 
from the atom with the lesser attraction. For example, if one atom of 
lithium, with a lone electron in its outer ring, which is rather loosely 
held, approached an atom of chlorine with seven electrons in its outer 
ring, rather firmly bound, the lithium atom would lose its electron to 
the chlorine atom, giving a positively charged lithium ion and a nega- 
tively charged chloride ion. 



These ions now have a structure similar to that of the rare-gas 
atoms and have no electron orbits in common. They therefore behave 
as individuals, are free to travel around, and are called ions (which is 
Greek for " wanderers"). Of course, they are somewhat restricted in 
their perambulations by their electrostatic charges, which cause them 
to attract oppositely charged ions and also neutral atoms. This type 
of linkage is the salt-forming, or heteropolar, or, simply, polar type of 
linkage. 

The Semipolar Bond. There is another type of bond of a different 
origin, which is somewhat intermediate in character between these 



32 QUALITATIVE ANALYSIS 

two types; this is the semipolar bond. We may again consider the 
case of two atoms approaching each other, but in this case, one of 
the atoms possesses a pair of unused electrons (called a lone pair) and 
the other needs two electrons to form a stable configuration. Here the 
atoms may share the pair of electrons (coming from one atom). A 
case of this type occurs in the reaction of ammonia and hydrogen ion. 
Ammonia has a lone pair of electrons, whereas hydrogen ion has none. 
They therefore share the pair of electrons, forming an ammonium 
ion. The whole ammonium ion now has the charge previously held by 
the hydrogen ion. 



H 



Lone pair 



H 
H:N:H 

ii 



Semipolar bonds also exist in neutral molecules, as in the product of the 
reaction of ammonia with boron trifluoride. 

H :F: H:F: 

H:N: + B:F: -* H:N:B:F*: 
H :F:" H : ^" 

Dipoles. In this type of molecule, one end has a greater negative 
charge than the other end, the molecule itself remaining neutral. A 
molecule of this type is said to possess a dipole. If molecules having 
dipoles are placed between two oppositely electrically charged plates, 
they arrange themselves so that their positive ends face the negative 
plate and their negative ends face the positive plate. This polarity 
of many molecules is important in explaining some of their properties. 

Valence. These three types of bonds are called the primary 
valence 1 forces that exist between atoms. Homopolar and semipolar 
bonds are called covalence, and heteropolar bonds are called electro- 
valence. For instance, the ammonium ion NH 4 + has four co valences 
and one electro valence; i.e., the hydrogen atoms are held by covalences, 
and the positive charge, the electrovalence, enables the ion to attract a 
negative ion. Similarly, carbon dioxide, C0 2 , has two covalent bonds; 
the carbonate ion CO 3 ~ has three covalent bonds and two electro- 
valences; hydrogen chloride, HC1, has one covalent bond; the hydro- 
nium ion HsO* has three covalent bonds and one electrovalence; the 
sulfate ion SOf has four covalent bonds and two electrovalences, etc. 

1 Secondary valence forces are the attraction due to van der Waals' forces. 



THE ATOM AND THE MOLECULE 33 

Valence Number. The valence-number concept is purely formal. 
It evolved independently (of the covalence, electrovalence idea) from 
the old valence theory. In this concept, certain key elements are 
assigned positive or negative numbers. The valence numbers of all 
other elements in molecules can then be calculated by addition and 
subtraction. For instance, in sodium carbonate, Na 2 CO 3 , ox/gen, O, is 
assigned a valence number of 2, sodium a valence number of +1, 
therefore 

[3(-2)] + [2( + l)] = -6 + 2= -4 

Since the molecule is neutral, the valence number of carbon is +4. 
Other examples are: 

Sodium sulfate, Na 2 SO 4 

[4(-2)] + [2( + l)] = -8 + 2 = -6 therefore sulfur = +6 

Potassium sulfite, K 2 S0 3 

[3(-2)] + [2( + l)] = -6 + 2 = -4 therefore sulfur = +4 

Sodium sulfide, Na 2 S 

Na = [2( + l)] = +2 therefore sulfur = -2 

Sodium thiosulfate, Na 2 S 2 3 
[3(-2)] + [2( + l)] = -6 + 2 ='-4 

therefore S 2 = +4 or S = +2 

Sodium permanganate, NaMn0 4 

[4(-2)l + ( + 1) = -8 + 1 = -7 therefore Mn = +7 

Hydrogen chloride, HC1 

II = +1 thcreforeCl = -1 

Methane, CH 4 

H 4 = 4( + l) = +4 therefore C = -4 



Chloromethanc, CH 3 C1 
(methyl chloride) 

[3(+l)I + [!(-!)] = +2 therefore C = -2 

Carbon tetrachloride, CC1 4 

C1 4 = 4(-l) = -4 therefore C = +4 

Chloroform, CHC1 3 

[3(-l)l + (+1) = ~2 therefore C = +2 

Dichlormethane, CH 2 C1 2 

[2(-l)] + [2(+l)] =-2 + 2 = therefore C = 



34 QUALITATIVE ANALYSIS 

The valence number of carbon in the last five compounds varies 
from +4 through to 4, but the number of covalent bonds is the 
same in all cases, four, and there are no electrovalences. This last 
example shows the artificiality of this concept, but it is nevertheless of 
some use. 

NOMENCLATURE 

Chemical compounds are named in a variety of ways. Compounds, 
which have been known since olden times have "trivial" names, which 
may or may not be commonly used at present. For instance, salt 
for sodium chloride, sal ammoniac for ammonium chloride, soda 
ash for anhydrous sodium carbonate, Glauber's salt for sodium sulfate 
decahydrate, Epsom salt for magnesium sulfate heptahydrate, etc., 
are names still in use. There are also other compounds whose " trivial " 
names arc not used at present to any great extent, such as aqua fortis 
for nitric acid, marine acid for hydrochloric acid (although muriatic 
acid is still used), white vitriol for zinc sulfate, (although blue vitriol is 
still used for copper sulfate), etc. 

Another type of name used is the common chemical name that may 
or may not be the same as the systematic name. For instance, the 
common chemical name for NaHC0 3 is sodium bicarbonate, and the 
systematic name is primary sodium carbonate. These common 
chemical names are those most usually employed in chemistry, 
especially in dealing with the simpler compounds. However, when the 
more complicated compounds are encountered, more systematic 
methods of nomenclature are employed. The principles used in 
naming the simpler compounds will be outlined here; the more complex 
compounds will be treated in the next chapter. 

Two-element compounds are named by using the name of the 
element having the positive valence number followed by the root of 
the name of the element having the negative valence number plus the 
ending -ide, as 

Calcium oxide CaO 

Lithium fluoride LiF 

Calcium chloride CaCl2 

Hydrogen chloride HC1 

Hydrogen sulfide H 2 S 

Aluminum chloride A1CU 

Lithium hydride LiH here H = 1 (experimentally) 

If the positive element has two valence numbers, the root of the 
name of the element with the lower positive valence number is given 
the suffix -ous, the higher, the suffix -ic, as 



THE ATOM AND THE MOLECULE 35 

Ferrous chloride FeCl 2 Fe = -f 2 

Ferric chloride FeCl 3 Fe - 4-3 

Cuprous chloride Cu 2 Cl 2 Cu = 4-1 

Cupric chloride CuCl 2 Cu 4-2 

Stannous chloride SnCl 2 Sn = 4-2 

Stannic chloride SnCl 4 Sn = +4 

Cerous oxide Ce 2 O 3 Ce = 4-3 

Ceric oxide Ce0 2 Ce = 4-4 

When two elements form a Aeries of compounds, the prefixes mono-, 
di-, tri-, etc., are added to the name of the element of negative valence 
number. This is usually done in the case of covalent compounds and 
less frequently in the case of clectrovalent compounds, as 

Titanium dichloride TiCl 2 Ti = 4-2 

Titanium trichloride TiCl 3 Ti 4-3 

Titanium tetrachloride Ti( -1 4 Ti = 4-4 

Sulfur monochloridc S-C1 2 S = 4-1 

Sulfur dichloride SC1 2 S = 4-2 

Sulfur tetrachloride SC1 4 S = +4 

Sometimes both systems are combined or mixed, as 

Nitrous oxide N 2 O N = 4-1 

Nitric oxide NO N = 4-2 

Nitrogen trioxide N 2 O 3 N = 4-3 

Nitrogen dioxide N0 2 N = 4-4 

Nitrogen tetroxide NO 4 N = 4-4 

Nitrogen pentoxicle N 2 O 6 N 4-5 

Arsenous chloride ) . , 

. . . . , , . , > As('l 3 

Arsenic trichloride ) 

Arsenic cliloride 
Arsenic pentachlorido 

Three-element compounds are named on the basis of their acids. 
To the root of the name of the negative radical is added the suffix -ous 
or -ic (to denote lower and higher valence numbers, respectively) and 
if required the prefix hypo-, meaning "less/' or per-, meaning "more." 
In the two-element acids, 1 the negative element takes the prefix hydro- 
and the suffix -ic, as 

HC1 Hydrochloric acid Cl - - 1 

HC10 Hypochlorous acid Cl . +1 

HC10 2 Chlorous acid Cl 4-3 

HC10 3 Chloric acid Cl - 4-5 

HC10 4 Perchloric acid Cl - 4-7 

H 2 S Hydrosulfuric acid S = 2 

H 2 S 2 O 4 Hyposulfurous acid S 4-3 

H 2 SO 8 Sulfurous acid S - 4-4 

H 2 S0 4 Sulfuric acid S 4-6 

1 See later chapter. Hydrochloric acid should really be written H a O + Cl~. 



36 QUALITATIVE ANALYSIS 

(Many of these acids may be unknown in a free state, but their salts 
are known.) To name the salts, the -ous suffix of the acid is replaced 
by -He and the -ic suffix by -ate; the word acid is omitted, and the name 
of the positive element is used before that of the negative element, as 

Na 2 S Sodium sulfide 1 

Na2S 2 O 4 Sodium hyposulfite 

Na2SO 3 Sodium sulfite 

Na 2 S0 4 ^ Sodium sulfate 

NaCl Sodium chloride 

NaCIO Sodium hypochlorite 

NaC10 2 Sodium chlorite 

NaClO 3 Sodium chlorate 

NaC104 Sodium perchlorate 

1 Two element compounds are, of course, named as in the preceding sections. 

If two positive elements replace the hydrogen in an acid, then both 
names are used, with the prefixes di- and tri- } if necessary. 

Na 2 KPO 4 Disodium potassium phosphate 

NaKC 4 H 4 O6 Sodium potassium tartrate 

If only one of the hydrogen ions in the acid is replaced by a positive 
ion, then the word primary urecedes the rest of the name; if two, the 
word secondary, etc. 

NaHCO 3 Primary sodium carbonate 

Na-jCOa Secondary sodium carbonate 

NaHSO 4 Primary sodium sulfate 

Na2SO 4 Secondary sodium sulfate 

NaH 2 PO 4 Primary sodium phosphate 

Na 2 HP0 4 Secondary sodium phosphate 

NaaPO 4 Tertiary sodium phosphate 

Ca(H 2 PO 4 ) 2 Primary calcium phosphate 

One other point must be considered. Oxygen acids are often 
formed by the reaction of the oxide with water. The number of 
molecules of water that are added to each molecule of oxide determine 
which acid will be formed. To illustrate this point, use will be made 
of the old Berzelius method of writing formulas. Here the prefixes 
meta-, pyro-, and orlho- give, in increasing order, the degrees of hydra- 
tion of the acid. 

P 2 O 6 -H 2 O B H 2 P 2 O 6 - 2HPO 8 . . . Metaphosphoric acid 

NaPOs... Sodium metaphosphate 

P 2 O 6 -2H 2 O H 4 P 2 O 7 Pyrophosphoric acid 

Na 2 H 2 P 2 O7 Secondary sodium pyrophosphate 

P 2 O6-3H 2 O H fl P 2 O 8 2H 8 P0 4 . Orthophosphoric acid 

Na 8 P0 4 . Tertiary sodium orthophosphate 



THE ATOM AND THE MOLECULE 37 

P 2 O 8 -H 2 O = H 2 P 2 O 4 2HPO 2 . . . Metaphosphorous acid 

NaP0 2 . . . Sodium metaphosphite 

P 2 O 8 -2H 2 O = H 4 P 2 O6 Pyrophosphorous acid 

Na 2 H 2 P 2 O6 Secondary sodium pyrophosphite 

P 2 3 -3H 2 ss H 6 P 2 O 6 s 2H 8 P0 8 . Orthophosphorous acid 

Na 8 P0 8 Tertiary sodium orthophosphite 

Unfortunately, inorganic nomenclature has inherited a mass of 
usages that tend somewhat to confuse the naming of compounds. 
There is little coordination in the naming of the compounds of many 
elements. For instance, in chloric acid, HC10 3 , an -ic acid, chlorine 
has a valence number of +5, whereas in sulfuric acid, H 2 SO 4 , another 
-ic acid, sulfur has a valence number of +6, and in stannic acid, 
H 2 Sn() 3 , tin has a valence number of +4. However, the rules given 
above hold in any given series of acids. 

READING REFERENCES 

LATIMER, W. M., and J. H. HILDEBRAND: " Reference Book of Inorganic Chem- 
istry," The Macmillan Company, New York, 1940. 

MARK, H.: "Physical Chemistry," Iriterscience Publishers, New York, 1940. 

PAULING, L.: "The Nature of the Chemical Bond," Cornell University Press, 
Ithaca, N. Y., 1939. 

QUESTIONS 

1. Draw a symbolic picture showing the whole electronic structure of fluorine, 
calcium, and potassium. 

2. Draw a symbolic, picture of the structure of ammonium sulfate, and name 
the various types of bonds. 

3. What is the valence number of carbon in the following compounds: CO 2 , 
CO, C 3 O 2 , C 4 H 9 C1, C 6 H 6 Br? 

4. Draw the symbolic picture of the structure of phosphirie, PH 3 . Would 
the compound be a dipole? If so, indicate the negative and positive portions of 
the molecule. 

5. Draw a symbolic picture of an actual compound that has a covalent bond ; 
an electrovalent bond; a semipolar bond. 

6. List the number of covalent arid electrovalent bonds in oach of the fol- 
lowing substances: IO 4 ~, PO 3 ~, P 2 O 7 , Cr 2 O 7 ~, SO 3 C1-, PC1 5 . 

7. Draw a picture of the electronic structure of the iron atom, showing all the 
electrons and indicating the valence electrons. 

8. Write the formulas and give the systematic names of all the acids that are 
derived from As 2 O 3 , arsenous oxide, and As 2 Os, arsenic oxide. 

9. Using the chlorine acids as an example, write the formulas and name all 
the hypothetical acids of fluorine. 

10. Given that KMnO 4 is potassium permanganate, what are the names of the 
following manganese compounds: Na 2 MnO 4 , NaKMnO 8 , HMn0 2 , H 2 Mn0 2 ? 

11. Name the following compounds systematically : K 2 H 2 Sb 2 O 7 , NaK(NH 4 )PO4, 
Na 2 HA10 3 , NaA10 2 , K 2 Se. 

12. Write the formulas of the following compounds: potassium perbromate, 
secondary sodium orthophosphite, arsenic pentaiodide, lead dioxide, silver cyanate. 



CHAPTER IV 
COMPLEX IONS AND MOLECULES 

Before we can predict, even in part, what the outcome of any 
reaction will be, we must possess an adequate knowledge of the 
participating components. We must, therefore, study not only the 
simple elements and compounds but also the more complex molecules, 
where the various atoms are linked by different types of bonds. Such 
molecules have traditionally been called complex ions and complex 
compounds. May we add, however, that the term complex does not 
necessarily mean complicated but is rather a generic term. 

Let us first consider an experiment that has been performed by 
many generations of students. When an excess of ammonium hydrox- 
ide is added to a solution of copper sulfate, the pale blue solution 
changes to a deep azure blue. Evidently a reaction has occurred. 
This reaction gives rise to two questions: (1) what has happened, and 
(2) can it be utilized? The practical man neglects the first question 
but investigates the possibilities of the second. The technologist, in 
his study of this second question, has discovered that such solutions 
have the property of dissolving cellulose and has used it to produce 
artificial silk. The analyst, noting that this intense and characteristic 
coloration invariably occurs whenever a soluble cupric salt is present, 
uses it as a simple tost. The scientist, however, prompted by his 
enlightened curiosity, looks for the answer to the first question: what 
has happened? By analysis, he finds that a compound has been 
formed that is composed of the cupric ion, four ammonia (NH 3 ) 
molecules, and whatever anion the solution originally contained. On 
further examination, the scientist finds that the four ammonia mole- 
cules are combined with the cupric ion, forming a new ion whose 
properties differ markedly from those of the original cupric ion. 

Occurrence. These peculiar types of ions, far from being rare in 
chemical reactions, are, on the contrary, so widely encountered that an 
understanding of their behavior provides a satisfactory explanation for 
many phenomena that disturbed and bewildered the earlier investi- 
gators. As an example, let us consider the separation of copper and 
cadmium ions, which is commonly used in analytical procedures. A 
solution containing both species of ions, if treated with hydrogen 

8 



COMPLEX IONS AND MOLECULES 39 

sulfide, would lead to the formation of both cupric sulfide, CuS, and 
cadmium sulfide, CdS. Of the two sulfides, cupric sulfide is very 
insoluble and would therefore precipitate before and more completely 
than cadmium sulfide. If the experiment is repeated, with the addi- 
tion of a slight excess of cyanide ion, CN~, before the saturation with 
hydrogen sulfide, a yellow precipitate of pure cadmium sulfide would be 
formed. 

This fact would appear utterly confusing if we did not know that 
both cadmium and copper ions form complex ions with cyanide ion. 
However, the cadmium complex is less stable than the copper complex 
and yields, in solution, a cadmium ion which combines with the sulfide 
ion, forming the characteristic yellow cadmium sulfide. 

Structure of Complex Ions. This experiment illustrates how com- 
plex ion formation can be used to make certain reactions understand- 
able. We now propose to examine critically some complex ions in 
order to gain a deeper insight into their structure or, to put it differ- 
ently, their atomic and electronic architecture. If the student con- 
siders tho complex ions mentioned thus far, he cannot deny being 
confused. His reasoning would run somewhat like this: I understand 
the mechanism that leads to the two positive charges on the cupric 
ion. The copper has lost two electrons to chlorine or some other 
element and therefore possesses two positive charges. In the ion 
Cu(NH 3 )4' } ", the ionic charge is the same as in the original ion. 
What factors, then, cause the four molecules of ammonia to attach 
themselves to the cupric ion? 

Whenever, in the study of natural phenomena, facts are encountered 
that do not fit into current patterns of thought, we can do one of two 
things: (I) ignore the recalcitrant facts or (2) insist on finding an 
explanation. The latter alternative often demands that we discard 
some of these patterns and replace them with newer models. 

Complex ions were largely ignored until the beginning of this 
century, when Alfred Werner singled them out and made them the 
object of a very exhaustive investigation. After studying all the 
available facts, Werner arrived at an interpretation of the structure of 
complex compounds. Let us follow Werner's reasoning with a definite 
example, hexamminocobaltic chloride, [Co(NH 3 )6]Cl 3 . The cobalt, 
says Werner, may be thought of as a sphere surrounded by a field of 
force. Let us assume that the six ammonia molecules are attracted to 
the " central atom" (cobalt, in this case) and since they arc subjected 
to exactly the same attraction will distribute themselves symmetrically 
around the cobalt. If we imagine the latter to be the center of an 
octahedron, each NH 3 would be located at one of the corners. The six 



40 



QUALITATIVE ANALYSIS 



molecules of ammonia are now held by the "central atom" in the 
first sphere of influence. The " central atom/' in addition to holding 
the six ammonia molecules in its first sphere, has a second sphere of 
influence wherein it can hold three .chloride ions. 



cr 




cr 



cr 



FIG. 18. Hexamminocobaltic chloride. 

We may ask at this point why Werner insisted on placing the six 
molecules of ammonia in the corners of a well-defined geometric figure? 
This question may be answered by saying that a good theory should 
provide more than a mere explanation. It should also enable us to 
predict the related phenomena. 



NH 3 



NH 3 i 



NH 3 




fCl 



Co 



Cl 



(a) 



NH 3 



(b) 



Fio. 19. Two isomers of the dichlorotetramminocobaltic ion. 

If we replace two of the six ammonia molecules with chloride ions 
(this has been done experimentally), we should obtain two compounds 
that differ, at least in some of their physical properties: solubility, 
color, etc. We should expect this because in Fig. 19a, the two halogens 
are close together, whereas in Fig. 196, they are as far apart as possible. 



COMPLEX IONS AND MOLECULES 41 

The actual experimental verification of this prediction emphasizes the 
importance of this theory. 

Coordination Number. The compound just studied has six ammo- 
nia molecules around the cobalt. Werner calls this the coordination 
number of the element and defines it as the number of molecules or 
ions held in the first sphere of influence. Upon examining numerous 
complex compounds, we find that 6 is by far the most frequently 
encountered coordination number. In our familiar example, 
Cu(NH 3 )4 ++ , copper has a coordination number of 4, whereas in 
Ag(NH 3 ) 2 + , silver has a coordination number of 2 and tungsten in 
W(CN) 8 a has a coordination number of 8. This suggests a reference 
to the periodic table and leads to the question, does the coordination 
number vary with the size of the atom and the atomic volume? It is 
not surprising, indeed, to find that carbon, nitrogen, and boron, i.e., 
elements of small atomic volume, never exceed a coordination number 
of 4, whereas molybdenum, to mention but one element of considerable 
size, exhibits a coordination number of 8 in K 5 Mo(CN) 8 . 

Theory and Hypothesis. Before proceeding to a modern elabora- 
tion of Werner's theory, it may be instructive to pause briefly to con- 
sider what is meant by "theory" and "hypothesis" and to consider 
their function and virtues in science. Facts may be compared to 
bricks. Bricks are useful in building a house, but a pile of bricks is not 
a house. Similarly, a mass of observations must be arranged into a 
pattern before it is called a hypothesis. The hypothesis becomes a 
theory upon growing older and proving its value in predicting new 
facts. It gains in prestige and gets promoted. Theories often reach 
such a venerable age that they outlive their usefulness, like the human 
executive who has lived to be a great asset to his firm but now is aged 
and is resting on his laurels. His past prestige, unfortunately, prevents 
a more vigorous and active person from taking his place. This sug- 
gests the function of theories in science. True, a theory coordinates a 
large number of facts and impresses a meaning on the pattern thus 
found, but its main purpose consists in stimulating research to discover 
new facts. A theory that has lost this tonic effect should be discarded. 
Werner's theory proved to be one of the most vivifying influences in 
inorganic chemistry. This should not, however, blind us to its limita- 
tions. We should like to know, for example, why six apparently neu- 
tral molecules were attracted to the metal ion in the first place? What 
mysterious force attracted them and held them at equal distances from 
the central atom? 

Modern Interpretation of Werner's Theory. In searching for the 
causes that bring about complex ion formation, let us consider first 



42 



QUALITATIVE ANALYSIS 



the finer structure of ammonia. 



We have previously seen that this 
H 



molecule may be represented as :N:H. We note that three of the 

H 

nitrogen valence electrons are utilized in forming covalent bonds, 
leaving a pair of electrons unengaged. Let us call this pair a lone pair 
of electrons. It should not be difficult to realize that, although the 
sum of the negative charges is counterbalanced by the same number of 
positive charges and hence the molecule as a whole is neutral, the dis- 
tribution of charges is nevertheless unsymmetrical. The result is to 
produce in the molecule a negative pole and a positive pole, the former 





FIG. 20. Octahedron inscribed in a cube. 

being centered in the region of the lone pair. Molecules of this type 
are called dipoles. 

Let us now consider how this concept is related to the structure of 
the hexamminonickelous ion Ni(NH 3 ) 6 ++ . Before doing this, we must 
examine the finer structure of the nickel ion. Nickel has an atomic 
number of 28, but since we are dealing with the nickelous ion that has 
lost two electrons from the N shell, we have, actually, 26 electrons, 8 of 
which are on the third sublevel of the M shell and hence find themselves 
at the outpost of the atom. Let us imagine that they occupy the 
corner positions of a cube. 

There is no necessity at all to assume that the electrons remain in a 
rigidly fixed position; it suffices to admit that the negative charge is 
concentrated at the corners of the cube. If we visualize the distribu- 
tion of electrical charges on a face of this cube, we can readily see that 
the more one moves away from a corner the less negative the charge 
becomes, until, upon reaching the center of the face, the charge 



COMPLEX IONS AND MOLECULES 



43 



becomes a minimum. In addition, we should remember that the 
positive charge of the atomic nucleus is exerting its maximum action 
at this same point. Since a cube has six faces, there arc six points of 
fractional positive charge. The six points are at equal distances from 
the center of the atom and, indeed, define an octahedron. We have 
thus returned to Werner's theory, but now we understand why the six 
dipoles (NH 3 in our example) would tend to attach themselves with 
their negative pole (the lone pair) to the six fractional positive regions 
around the central atom. 

This can be illustrated by a structural formula in which, for the 
sake of clarity, only the strategic electrons have been drawn (Fig. 21). 



Fia. 21. Symbolic representation of the hexamminonickelous ion. 

It is apparent that in such a case each coordinated molecule shares its 
lone pair of electrons with the central ion. In the Sidgwick concept, 
each coordinating molecule " donates" its lone pair of electrons to the 
central ion. This sharing creates a bond, variously designated as 
semipolar bond or coordinate link, and is often represented by an arrow 
pointing from the electron pair to the receiving central ion. This 
concept is a help to the imagination and therefore useful, but the 
student should not be led to a belief in the actual existence of one-way 
traffic signs in the make-up of complex ions. 

Types of Complexes. In the light of this explanation, we should 
now investigate the often expressed idea that " inorganic chemistry 
deals with a comparatively small number of well-defined compounds, 
most of which have been exhaustive!}' studied." The fallacy of this 
view is easily demonstrated. Let us select a single ion, e.g., Co* 4 " 4 *, 
having a coordination number of 6 and thus having six positions to be 
filled with dipole molecules such as NH 3 , pyridine, water, NO, etc., or 
with anions such as halide, N02~~, CN~, etc. We proceed now to 
combinations, e.g. 



44 



QUALITATIVE ANALYSIS 



NH 8 H 2 


+ 


N0 2 


= 


Br 




H 2 






N0 2 Cl 




N0 2 Cl 




NH 3 H 2 


NH 3 Co Cl 




Oo 




Co 




Co 






N0 2 H 2 




NH 3 N0 2 




CN H 2 


NH 3 N0 2 




C) 




NH 3 




CN 



NO 

NH 3 CN 

Co 
NH 3 NH 3 

NO 



Cl 

NH 3 NH 3 

Co 
Cl NH 3 

Cl 



Cl 
Cl Cl 

Co 
Cl Cl 

Cl 



We can readily see that the number of possible complex ions and com- 
plex compounds becomes quite large. Consider, too, that thus far we 
have confined ourselves solely to cobalt compounds. If we varied the 
coordinating ion as well as the groups of molecules and ions, we should 
produce practically an unlimited number of compounds. The suc- 
cessive replacement of dipole molecules such as H 2 O by anions such as 
Cl~ deserves further comment. 

Let us consider chromic chloride hexahydrate, a compound char- 
acterized by a beautiful violet color due to the complex hexaquo- 
chromic ion Cr(H 2 0) 6 +++ . We could easily determine that the solu- 
tion contains three chloride ions for each complex ion. This we could 
do, for example, by adding silver nitrate solution and weighing the 
silver chloride precipitate formed. Let us now exchange a Cl~ for a 
molecule of H 2 O, thus producing the chloropentaquochromic ion 
[Cr(H 2 O)5Cl]++. (The practical details of the preparative procedure 
need not concern us here.) We see first that the charge of the complex 
ion has dropped from plus 3 to plus 2. The explanation is readily 
forthcoming. The incorporation of a negative ion in the first sphere of 
influence uses one of the electron pairs of the ion, thus forming a 
coordinate link. However, the negative charge on the chloride ion 
remains. Therefore, the total charge on the ion is composed of plus 3 
from the chromium and negative 1 from the chloride, leaving a net 
charge of plus 2. This change has definitely altered the complex ion. 
The color of the solution now is deep green. By adding a solution of 
AgN0 3 to this solution, we could prove the presence of 2 Cl~ for each 
complex ion. A further exchange of Cl~ for H 2 O would yield the 
dichlorotetraquochromic chloride [Cr(H 2 0)4Cl 2 ]Cl, a light green 
compound. 

As a further illustration, let us use hexamminocobaltic chloride 
and successively replace the ammonia molecules with the monovalent 



COMPLEX IONS AND MOLECULES 



45 



nitrite ion. The numerical value under each formula represents the 
conductivity of a 0.002JV solution of each compound. 



[Co(NH 3 ) 6 ] +++ 3C1- 

412 
[Co(NH 3 ) 4 (N0 2 ) 2 ]+ Cl- 

97 
[Co(NH 3 ) 2 (N0 2 ) 4 ]- K + 

97 



[Co(NH 3 ) 6 (N0 2 )]++ 2C1- 
240 

[Co(NHi),(NOOiJ 
1.5 

[Co(N0 2 )a] 3 3K+ 
418 



Note that the conductivity decreases sharply with the decrease in 
positive charge. The compound Co(NH3) 3 (N() 2 ) 3 is neutral. Com- 
plexes of this type are soluble in organic solvents such as chloroform, 
carbon tetrachloridc, etc. Further exchange of ammonia and nitrite 



H 2 
H 2 



Cr 



Hexaqyochromic Ion 



H 2 
Hjp H 2 

Cr 
H 2 H 2 

CL 



Chloropentaquochromh Ion 



H 2 



fCl 



H 2 



Wr 



7 ci 



Cr 



DichloroMraauochromic 
(two -forms) 
FIG. 22. 



Ion 



ions imparts a negative charge and transforms the complex into a 
complex anion. Consequently the conductivity increases and reaches 
a maximum with the hexanitritocobaltate ion [Co(NO 2 )6] s , the end 
product of our scries of substitutions. 

Stability of Complexes. li is to be expected that the stability of 
complex ions and complex compounds will vary considerably. 
Co(NH 3 ) 5 + + + is remarkably stable, whereas the corresponding cobalt- 
ous compound Co(XH3)6 ++ dissociates to a considerable extent in 



46 QUALITATIVE ANALYSIS 

solution. This may appear surprising and suggests the question, is it 
possible to predict, at least qualitatively, whether a given complex 
ion will prove stable or not? Such predictions are made possible 
by the hypothesis of the effective atomic number, E.A.N. We recall 
that "atomic number" Defers to the number of positive charges of an 
atomic nucleus, which number also coincides with the number of elec- 
. trons surrounding such a nucleus. "E.A.N." refers to the total num- 
ber of electrons surrounding the central atom of the complex. This is 
best illustrated with a simple example. Let us avssume'that we wish to 
determine the effective atomic number of copper in Cu(NH 3 ) 4 ++ . 
The atomic number of copper is 29, which tells us that 29 electrons 
surround the nucleus in the neutral copper atom. In the cupric ion, 
however, the electrons belonging to copper are 29 2 = 27. To this 
must be added 8 electrons emanating from the 4 electron pairs directly 
coordinated in making up the complex. Therefore the E.A.N. of the 
complex ion is 27 + 8 == 35. 

The hypothesis now states that whenever the effective atomic num- 
ber of a complex ion coincides with the atomic number of a rare gas, we 
can expeet greater stability than when it does not. As an example, let 
us choose the ferrocyanide ion, a well-known stable complex. Its 
systematic name is hexacyanohypoferrite ion. The effective atomic 
number of the iron in the compound equals 36. (Atomic number 
of Fe = 26 2 = 24 for the atomic number of ferrous ion +12 for 
the six CN groups = 36.) The rare gas krypton has an atomic num- 
ber of 36. It follows that the ferrocyanide complex should be found 
stable, a fact that can be demonstrated by purely experimental 
methods. Now let us take an unstable complex, the hexammino- 
cobaltous ion, mentioned previously. The effective atomic number of 
cobalt in this complex is 37, which does not coincide with a rare-gas 
atomic number. We should therefore predict instability, which again 
is actually found experimentally. However, hexamminocobaltic 
ion is stable, since the effective atomic number of cobalt is 36, coincid- 
ing with the atomic number of krypton. 

Thus far we have concerned ourselves with the general aspects of 
the complex-ion problem, their structure, formation, and stability. 
Our next task is comparable to that of Ali Baba, who found himself in 
the treasure cave amidst unending rows of gold pieces and jewels. He 
was faced with the difficult task of choosing which of the treasures to 
carry off with him. In attempting to record and describe some inter- 
esting examples of complex ions, we must choose from a wealth of 
material of which only a specialist can hope to keep track. It seems 
wise, therefore, under the circumstances, to refer the more inquisitive 
student to some excellent treatises on the subject (a list of which will 



COMPLEX IONS AND MOLECULES 47 

be found at the end of this chapter) and otherwise to content ourselves 
with recording a few examples of particular interest to the analyst. 

Ammonia Complexes. To this class belong many compounds used 
in qualitative analysis either for identification purposes, as the intense 
blue Cu(NH3)4 ++ (tetrammino cupric ion) or for separations, as 
Ag(NH 3 )2 + (diamminoargentic ion). The latter is formed whenever 
silver chloride is treated with ammonia, the silver going into solution 
and so being removed from any other insoluble chlorides present. A 
study of the formulas of the complex ions and molecules already 
mentioned will show that it is the lone electron pair in the ammonia 
molecule that binds it in complex ions. From this we could predict 
that the other three electrons need not necessarily be shared with 
hydrogen atoms but that other atoms or groups of atoms could be 
substituted for hydrogen. 

H R R R 

:N:H :N:H :N:R :N:R 

ii ii ii it 

Ammonia Monosubsti tuted Disubstitutcd Trisubstituted 

ammonia ammonia ammonia 

There is a great deal of experimental evidence that this is the case. 
Consider the formula of pyridine, a 
liquid of unpleasant odor but of great 
usefulness. It may be conceived as . 

being derived from ammonia, in which // . * 

the three electrons are shared with C . 

carbon atoms, leaving the lone pair of 
nitrogen intact. 

Pyridine, therefore, may be sub- 
stituted for NH 3 in complex ions. 

Best known in qualitative analysis are C ^ c 

the "pyridonium" complexes of zinc, 
cadmium, and cobalt. Pyrrole, which 
is a close relative of pyridine, as can Y 

be seen from its formula, leads us to _ 00 " one pair 

,.,.,,_ , FIG. 23. Pyridine. 

the question, which is the most 

important chemical compound in the entire realm of nature? 

There can be no hesitation about the answer; it is chlorophyll, the 
green pigment of leaves and grass, a coordination complex in which 
the central atom, magnesium, is surrounded by four pyrrole nuclei. 
Is it a mere coincidence that we should encounter complexes so close 
to the roots of life? 

Hydrates and Aquo Complexes. In one of the preceding para- 
ffranhs. w fino.ountfirfid tthromin chloride hexahvdrate. The fact that 



48 QUALITATIVE ANALYSIS 

molecules of water function very much like NH 3 molecules as con- 
stituents of complex ions is so widely encountered that we often ignore 
the coordinated water molecules. The three forms of chromic chloride 
demonstrate, however, that the replacement of a single H 2 in the 
chromic complex with a chloride ion (as previously discussed) brings 
about a visible marked change in the properties of the compound. 



HC - CH .^ ... 

' ' CIH 



\ 



' N 

; 

Lone pair? 



H 

Fia. 24. Pyrrole. 

The original ion Cr(H 2 O) 6 +++ was violet, whereas the resulting 
[Cr(H 2 O)5Cl]" 1 " 1 " (pentaquochlorochromic ion) is green. Molecules of 
H 2 0, as We can see, can be constituents of anionic as well as cationic 
complexes. Whenever entire water molecules arc coordinated into 
complex ions, we speak of hydrates or aquo complexes. We know, 
however, of instances, by no means rare, in which only the hydroxyl 
ion is incorporated into cationic complexes, as, for instance, in the 
pentaquohydroxozinc ion [Zn(H 2 O)5(OH)] + . This ion is found in zinc 
chloride solutions and accounts for the peculiar properties of such a 
solution (which will be discussed in greater detail in Chap. VII). 

Acido Complexes. The reader may not be aware of it, but the 
fact remains that he is already quite familiar with some compounds 
of this class. Let us recall the successive replacement of ammonia 
molecules with nitrite ions, starting with Co(NH 3 )6 + " l " + . From the 
fourth replacement on, the original cation has been changed into an 
anion, yielding as the final product Co(NO 2 ) 6 -. This ion immediately 
brings to mind a large number of similarly constituted anions that 
are often encountered in analytical chemistry. Best known are 
probably the ferro- and ferricyanide ions, Fe(CN) 6 "^ and Fe(CN) 6 s , 
which not only are used as reagents for the detection of iron but are 
also of great industrial importance in the production of paint pigments. 
The formula of the mineral cryolite, indispensable in the production of 
aluminum, is often written as AlF 3 -3NaF. Such "double-salt" 
formulations should be viewed with suspicion, since they are, as a rule, 
leftovers from the past, when complex ions were not properly under- 
stood. The correct formula for cryolite should, of course, be Na 3 AlF 6 , 
and its name is trisodium hexafluoraluminate. 



COMPLEX IONS AND MOLECULES 49 

The hexafluoroferrite ion, FeF 6 - is a good example of a practical 
application of complex ion formation to analytical procedures. Let us 
suppose that a certain test is interfered with by the presence of ferric 
ion. We could, of course, devise a method of converting the iron into 
some insoluble compound and then filter it off. This, however, would 
be time-consuming and probably would result in incomplete removal. 
The addition of fluoride ion offers a much more elegant solution to our 
problem. The ion would be incorporated into a very stable complex 
and thus would not interfere with the test. 

As our last example, we choose Nessler's reagent. This reagent 
is indispensable in water analysis in that it enables one to detect 
minute amounts of ammonia (which, in turn, indicate decaying organic 
matter). Nesslcr's reagent is prepared according to the following 
equations: 

4K+ + 41- + HgCl 2 -> 4K+ + Hglr + 201- 



It reacts as follows: 




N 



+ 30H-+:N:H- 

N 



H 
:N:Hg:6:Hg:I: 



Brown precipitate 



Polyhalides, Polysulfides, and Polyacids. The student has now 
acquired the firm conviction that he has gathered all the essential 
facts pertaining to complex ions and that the remainder of this chapter 
is merely an elaboration of some details with suitable examples. 
Nature, however, seems to delight in disappointing the investigator 
who thinks that he has at last succeeded in smartly classifying a certain 
set of phenomena. Surrounding the central atom in the first sphere of 
influence, we have thus far encountered molecules such as NHs, or 
H 2 O, dipoles, in other words. We also found ions such as Cl~ or CN~, 
etc. We now proceed to molecules. It is well known that only 
minute amounts of iodine dissolve in water. The dark crystals of the 
element, however, easily dissolve in a potassium iodide solution. The 
generally accepted explanation assumes the formation of the (I-I2)"" 
complex. This symbolic representation implies that the iodide ion 
acts as the central atom binding the molecule of iodine. If, instead of 
KI, we select Csl solution, a still larger amount of iodine can be dis- 
solved, and the ions found are formulated as 



Cs+ :I: 




60 QUALITATIVE ANALYSIS 

Ammonium polysulfide is another example of such elemental complexes, 
as we may properly call them. This reagent has the remarkable 
property of dissolving the sulfides of arsenic, antimony, and tin and is 
therefore widely used in analytical schemes to effect the separation of 
the Group II sulfides into subgroups. The student memorizes the 
formula (NHU^S* for ammonium polysulfide and seldom gives any 
thought to the value of x. It should, however, be written as 
(NH^S'Sy, wherein y may vary from 1 to 4. 

Inner Complexes or Chelate Compounds. We would predict 
without hesitation that whenever NH 3 is added to a solution containing 
Cu ++ ion, the complex Cu(NH 3 )4 + + will form. We now propose to 
modify slightly this familiar experiment. Instead of cupric ion, sup- 
pose that we were dealing with a water-soluble but un-ionized copper 
compound. Would copper retain its ability to coordinate ammonia 
molecules? We should find by experiment that it would. Now sup- 
pose that the copper compound had somewhere in its molecule an 
amino group, NH 2 . These conditions would be realized in the follow- 
ing compound: 

O Bf 

II II -:C:H 

CH 2 C O Cu C CH 2 H : N : H 

NH 2 NH 2 

We see that the amino group NH 2 has retained the electron dis- 
tribution of the ammonia molecule. There remains, in particular, the 
lone electron pair, the indispensable requirement for complex forma- 
tion. The experiment reveals that copper glycinate forms, indeed, a 
complex compound, characterized by an intense blue color. The 
structure of the compound could be represented as follows: 




FIG. 26. Copper glycinate. 



COMPLEX IONS AND MOLECULES 51 

It is significant that since both the central metallic atom and the 
coordinated groups are parts of the same molecule, the complex forma- 
tion necessitates a "rolling up" or cyclization of the molecule. In our 
example, two five-membered rings have been formed. The designa- 
tion "inner complexes" or chelate (i.e. y clawlike) compounds seems 
therefore quite appropriate. The application of most of the organic 
reagents described in our experimental part is founded on their ability 
to form such chelate compounds. A typical example is nickel dimethyl 
glyoxime, which precipitates as a beautiful red compound whenever 
dimethyl glyoxime is added to an ammoniacal nickel solution. The 
formula is 

CH 3 CH 3 

I i H ' c CTT ' 
C C c . ('i 

II II .. ' ".. 
ON N OH H:O:N N:O: 



X 

-N 

L- 

CH 3 CH, 



or :NJ: 

:0:N N:6:H 
'C : C 

H 3 C . CH 3 



APPLICATION OF COMPLEX IONS TO ANALYTICAL PROCEDURES 

A certain insight into the general types of complex ions and complex 
compounds having been gained, it seems appropriate to group such 
complex ions that are of special interest to the analytical chemist and 
consider them separately. 

1. The extensive use of complex ions and compounds in final 
identification tests should be stressed. The formation of the intensely 
colored Cu(NH 3 )4" l " + ion from ammonia and the Cu^^O)^ 4 " often 
suffices definitely to establish the presence of copper. The ferro- 
cyanide ion, Fe(CN) 6 ^, is used as a sensitive reagent for ferric ion, since 
the two form a deep blue precipitate, Prussian Blue, Fe 4 [Fe(CN) 6 ]3. 
The frequent use in recent years of organic reagents like diphcnyl 
carbazide, rhodanine, thiosinamine, etc., as confirmatory tests is based 
on their forming intensely colored complex compounds of the chelate 
type with particular cations. The distinctive advantages of these 
reagents are their sensitivity and their specificity. The chelate com- 
pounds obtained by these reactions are very often of the nonionic type 
and hence are easily soluble in organic solvents. This enables us to 
increase the sensitivity of many tests, since we are thus able to con- 
centrate the final identification product by extraction with a small 



52 QUALITATIVE ANALYSIS 

volume of an organic solvent. A good example of this is mercuric 
diphenylcarbazone which can be extracted with a small amount of 
amyl alcohol to give an intensely blue solution. 

2. Complex formation is often utilized in various separation pro- 
cedures. As a simple example, let us consider the separation of 
silver chloride and mercurous chloride in the first analytical group. 
The addition of an ammonia solution to this mixture will result in the 
solution of the silver chloride because of the formation of the diam- 
minosilver ion, Ag(NH 3 ) 2 +. The mercurous chloride will at the same 
time be converted to an insoluble mixture of mercury and mercuric 
amido chloride, HgNH 2 Cl. Another example is the separation of the 
sulfides of Group II into two subgroups. The sulfides of arsenic, 
antimony, and tin are soluble in warm ammonium polysulfide solution, 
forming the thio complexes; the thioarsenate, AsS 4 s , the thioanti- 
monate, SbS^, and the thiostarmate, SnSa 553 , respectively; the sulfides 
of the other metals in this group remain practically unaffected. 

Another important analytical procedure is the separation of 
cadmium from copper. If a mixture containing the tetramminocupric 
ion and thfe tetramminocadmium ion is treated carefully with cyanide 
ion, the following series of reactions occurs : 

2Cu(NH 3 ) 4 ++ + 4CN- - 2Cu(CN) 2 + 8NH 3 

2OH- + 2Cu(CN) 2 - Cu,(CN), + CNO~ + H 2 O + CN~ 

white 

Cu 2 (CN) 2 + 4CN- - 2Cu(CN) 3 = 
Cd(NH 3 ) 4 ++ + 2CN- -> Cd(CN) 2 + 4NH 3 
Cd(CN) 2 + 2CN- - Cd(CN) 4 - 

The cyanocadmium complex is less stable than the cyanocuprous 
complex; hence when hydrogen sulfide is added, pure cadmium sulfide 
is precipitated. (For a discussion of the quantitative aspects, see 
Chap. VIII.) 

Lastly, it should be pointed out that complex-ion formation is not 
invariably a blessing as far as the analyst is concerned. Suppose, to 
give only one well-known example, that a solution is to be analyzed for 
aluminum, iron, and chromium. The addition of a slight excess of an 
ammonia solution would ordinarily suffice to precipitate these three 
metals as the hydroxides. However, in the presence of many organic 
acids, such as citric and tartaric acids, the precipitation is partly or 
completely inhibited because of the formation of citrate or tartrate 
complexes. This problem can, of course, be solved, but the solution 
involves the complete destruction of the offending acids by a rather 
tedious procedure. 



COMPLEX IONS AND MOLECULES 53 

OUTLINE OF THE SYSTEMATIC NOMENCLATURE 
OF COMPLEX COMPOUNDS 

1. The name of the cation precedes that of the anion. 

2. In naming the coordination complex, the groups are named in the 
following order: 

NOTE: If the name of a coordinated group ends in -idcj this ending is replaced 
by an -o; if the name ends in -ite or -ate, the final -e is replaced by an -o. 



a. Acid groups: Cl~, chloro; Br~, bromo; SC^", sulfato; 

carbonato; CN~~, cyano; CNO", cyanato; N02", nitrito; etc. 
6. Neutral groups containing oxygen: H 2 O, aquo; O~,*oxo; O2~, 

peroxo; OH~, hydroxo; etc. 

c. Ammonia groups : NH 3 , ammino. 

d. Miscellaneous groups: S~, thio; NH 2 CH 2 r>H 2 NH2, ethylene 
diamino; etc. 

NOTK: Each of the preceding groups has a prefix (if more than one is present) 
denoting the number of such groups in the ion. The prefixes used are di-, tri-, 
tetrar, penta, hexa-, etc. 

e. The name of the central or coordinating atom with the appro- 
priate ending. If the ion is a cation, it has the same ending 
that the unsubstituted ion would bear. If the ion is an anion, 
it has the ending of the corresponding anion. 

Examples: 

[Co(NH3) 6 ]Cl 3 ........... Hexamminocobaltic chloride 

[Cl(H 2 0)(NH 3 ) 4 Co]Cl 2 . . . . Chloraquotetramminocobaltic chloride 

K[Co(NH 3 ) 2 (NO 2 ) 4 ] ....... Potassium tetranitritodiamminocobaltate 

K 3 [RhCl 6 (H 2 O) 3 ] ......... Tripotassium hexachlorotriaquorhodite 

K[AuBr4(H 2 0) 2 ] .......... Potassium tetrabromodiaquoaurate 

READING REFERENCES 

EMELEUS, H. J., and J. S. ANDERSON: "Modern Aspects of Inorganic Chemistry," 
1st ed., D. Van Nostrand Company, Inc., New York, 1940. 

MORGAN, GILBERT T., and FRANCIS H. BURSTALL: "In6rganic Chemistry," 
W. Heffer & Sons, Ltd., Cambridge, England, 1936. 

THORNS, P. C. F., and A. M. WARD: "Ephraim's Inorganic Chemistry," 3d ed., 
Gurney and Jackson, London, 1940. 

QUESTIONS 

1. What is the coordination number of the copper-ammonia complex formed 
in analytical Group IIA? Of the silver-ammonia complex in Group I? Of the 
zinc-ammonia complex in Group III? 



54 QUALITATIVE ANALYSIS 

2. What is the coordination number of iron in potassium ferrocyanide, 
potassium ferricyanide, sodium nitroprusside? 

3. What is the E.A.N. of nickel in [(NH 3 ) 2 (H 2 O) 2 (NO 2 ) 2 Ni]+? In [(Nil,),- 
(H 2 0) 2 (N0 2 )Ni]++? 

4. Write the formulas and names of the ions produced by the successive 
replacement of NH 3 by Cl~ in the hexamminochromic ion. 

5. In making sodium nitroprusside from sodium ferricyanide, how do you 
explain the change in the charge of the anion? 

6. Draw pictures of all the possible isomeric forms of the PtF 4 Cl 2 "" ion; of the 
PtF 3 Cl 3 - ion. 

7. Name the following compounds: NiSiF 6 , [(H a O)j(NH 8 )4Ni](NOj) s , 
Na 3 AsS 3 0, Na 3 SbS 4 . 

8. Write the formula of the following compounds: potassium triiodoplumbite, 
sodium dibromodinitritodiaquoaunite, dipotassium peiitanitritoamminocobaltate. 

9. Give the names of the following ions: FeF 6 " ; Cr(SO 4 ) 3 =s , Cu(CN) 3 = . 

10. Write the formulas of the following compounds: hexafluorosilicic acid, silver 
dicyanoargentate, trinitritodiaquoarnm inocobalt. 



CHAPTER V 
THE KINETIC THEORY OF MATTER 

We have all at some time made the observation that substances 
exist in three physical states: gas, solid and liquid. The question 
now before us is, what is the correlation between the known facts of 
atomic and molecular structure and these three states of matter? 

In the preceding chapters, we have considered single atoms and 
molecules, whereas actual substances with which ^e are familiar are 
composed of very large numbers of molecules. We must therefore 
study the effects produced when this condition exists. 

Water, for example, can exist in the solid, liquid, or gaseous state, 
and in each of these states the molcculo H 2 O remains unchanged. The 
factors responsible for the physical differences must therefore lie in the 
forces that act among these kirge numbers of molecules. We know 
that the gas, steam, can be quite highly compressed, and since it is not 
probable that the molecules are themselves compressed, it seems 
plausible to say that there is a large amount of free space between 
molecules. Liquids, such as water, and solids, such as ice, are, how- 
ever, compressible only to a slight extent, and we may therefore con- 
clude that in these latter states the molecules are closer together. 

In order to gain a deeper insight into the interplay of forces existing 
among large numbers of molecules, we must consider these states of 
matter separately. The simplest case is that of the gases, and we shall 
therefore study this state first. 

GASES 

How may a gas be characterized experimentally? One of the 
criteria of a gas is its lack of shape; i.e., it fills completely any 
container into which it is placed. The forces of gravity have no 
apparent effect on gases. 1 This point may be illustrated by the 
ingenious experiment performed by Dalton in 1801. He connected 
two bottles, one filled with hydrogen and the other with carbon 

1 The inquisitive student may ask why, if this statement is true, does not the 
atmosphere surrounding the earth float away. The answer is that gravity does 
affect gases but to such a small extent that it is unnoticeable as long as experi- 
mentally practical quantities are used. 

55 



56 



QUALITATIVE ANALYSIS 



dioxide, with the bottle of hydrogen at the top (Fig. 26). After some 
time, he removed the connecting tube, stoppered the bottles, and 
after inverting the bottles in a solution of potassium hydroxide (which 
absorbs carbon dioxide), removed the stoppers. The liquid rose to the 
same level in both bottles. This can be explained only if the gases have 
mixed spontaneously and uniformly. This, in turn, must mean that 
the molecules of gases are in continuous motion and since this condition 
of equilibrium was attained with fair rapidity, the molecules must be 
moving at high velocity. 



KOH ' Solution \ 







FIG. 26. Dalton's experiment. 

Another criterion of gases is their low density when compared 
to liquids and solids. For example, 30,600 ml. of water vapor at 
100C. has approximately the same weight as 18 ml. of water at 
100C. A third criterion that we have mentioned previously is high 
compressibility. All these experimental observations and the phe- 
nomena attendant upon them may be explained in terms of the kinetic 
molecular theory. 

The Kinetic Theory. According to this theory, gases are composed 
of a swarm of molecules, comparatively far apart, each molecule mov- 
ing independently of the others at high velocity and in a straight line. 
The energy of this motion is called the kinetic energy of the gas and 
depends directly upon the number of molecules, the mass of each, and 
their average velocity. Since the molecules move in a straight line, it 
is evident that their paths will change only when they "collide" with 
other molecules. 1 Or, since the molecules move independently of one 

1 By collision we mean not actual collision but rather that the molecules swerve 
when they approach one another very closely (see Chap. VI). 



THE KINETIC THEORY OF MATTER 57 

another in a straight line, the forces of attraction and repulsion inherent 
in each molecule evidently have no effect upon each other. This is 
explained by assuming that the molecules are generally very far apart 
and that the forces at these distances are too weak to exert any influ- 
ence. The only exception occurs when molecules undergo collision. 

This qualitative picture has been expressed mathematically, and 
as a result, we are able to make certain useful deductions. We may, 
for example, calculate the average or mean velocity of a gas molecule. 
For hydrogen, this is 183,000 cm. per sec., and for oxygen, the value 
is 46,000 cm. per sec. at S.T.P. It is obvious that each molecule 
undergoes innumerable collisions in this time. As a matter of fact, 
the average distance that a molecule travels before it collides with 
another is about 0.00001 cm. at standard conditions. This value is 
called the mean free path of the gas. The average distance between 
molecules at 0C. and 1 atm. pressure is about thirteen times thoir 
diameter. The number of molecules in any gas may be calculated 
from the well-known Avogadro number, which states that, at stand- 
ard conditions 22.4 liters of a gas contain 6.06 X 10' 23 molecules or 
2.7 X 10 22 molecules per liter. 

GAS LAWS AND THE KINETIC THEORY 

Boyle's Law. Many years before the kinetic theory of gases was 
postulated, certain regularities in the behavior of gases were noted. 
Those were eventually formulated into the gas laws. The oldest of 
these is Boyle's law, which states that, at constant temperature, the 
pressure exerted by a gas is inversely proportional to the volume it 
occupies, or that the pressure multiplied by the volume of the gas is a 
constant. This may be expressed mathematically by 

P ~ y or PV = k 

The question may be asked, how may this law be correlated with 
the kinetic theory? The first point to be settled is to explain pressure 
on the basis of the kinetic theory. Since most gases are kept in con- 
tainers, the molecules of the gas, in addition to colliding with each 
other, collide with the walls of the vessel. In fact, the walls of the 
vessel are subjected to a continual bombardment of molecules, very 
much like hailstones striking a roof. This gives rise to a pressure on 
the walls of the container. The greater the number of molecules 
striking the wall the greater is the pressure exerted on the wall * It is 
obvious that if the same number of molecules occupied half the volume, 
there would be twice as many collisions with the walls and conse- 



58 QUALITATIVE ANALYSIS 

quently the pressure would be twice as great. For example, 4 grams of 
helium occupy a volume of 22.4 liters at 0C. and 1 atm. pressure, but if 
the volume is changed to 1L2 liters, then the pressure becomes 2 atm. 
However, the product of the pressure and the volume remains constant 
at 22.4 liter-atm. If the volume were reduced to one-quarter of the 
original volume (5.6 liters), then the pressure would be four times as 
great (4 atm.), and the pressure-volume factor would still be constant. 
This is a simple explanation of the law that was first enunciated by 
Robert Boyle in 1662 as a result of his researches. 

Charles's Law. The next gas law to be considered is Charles's 
law, which states that the volume occupied by a gas at constant pres- 
sure is directly proportional to its absolute temperature, or, stated 
mathematically 

V ~ T or ~ = k 

Since, from Boyle's law, we know that P ~ 1/7, hence P and 7 are 
interchangeable, and wo may write P ~ T. This means that the pres- 
sure depends upon some quantity called the temperature. What is the 
meaning of temperature when the term is applied to groups of mole- 
cules? Or, to phrase this question differently, what conditions other 
than change in volume will cause an increase in the number of collisions 
with the walls? If a greater velocity is imparted to the molecules, 
they will strike the walls more frequently and consequently cause an 
increase in pressure. This, ultimately, is the meaning of temperature 
as applied to gases. When a gas is heated, the molecules move more 
rapidly, and as a result, there is an increase in the pressure if 
the gas volume is kept constant, or, if the pressure is kept constant, the 
volume increases. 

Dalton's Law. This brings us to Dalton's law of partial pressures, 
which states that, in a mixture of gases that do not react with each 
other, each gas exerts its pressure independently of the other gases, as if 
it alone were filling the vessel. In a mixture of two gases A and B in a 
vessel, the total pressure exerted is equal to the total number of col- 
lisions with the walls. The pressure exerted by gas A will be due to 
the collisions of molecules of A with the walls, regardless of the presence 
of gas B. If the molecules of A undergo one million collisions with the 
wall in a given unit of time, the total pressure exerted by A will be the 
pressure due to the million collisions. If molecules of gas B are intro- 
duced and if B does not react with A, the total number of collisions 
will increase, causing an increase in the total pressure. However, the 



THE KINETIC THEORY OF MATTER 59 

pressure exerted by A remains the same, whereas the total pressure 
exerted is the sum of the pressure of A plus the pressure of B. 

The Combined Gas Laws. For practical purposes, it is often 
advantageous to use the combined form of the Charles's and Boyle's 
law. This is customarily formulated as 

PV = nRT 

where P = pressure 

V = volume 

T = absolute temperature 

R = universal gas constant for 1 mole 

n number of moles of gas 

Experimental Verification of the Gas Laws. The gas laws are 
expressions of experimental facts. However, the practical evidence 
upon which they were based has been found to be not quite so exact as 
the laws would have us believe. For instance, according to Charles's 
law, gases at constant pressure should expand ^73 of their volume for 
each degree rise in temperature. Careful experiments have shown that 
all gases do not exhibit this same rate of expansion. Table XIV lists 
the rates of expansion of several gases with increasing temperature at 
constant pressure. 

TABLE XIV. RATE OF EXPANSION OF GASES PER DEGREE RISE IN TEMPERATURE 

AT CONSTANT PRESSURE 

Air 1/272.8 Oxygen 1/272.2 

Hydrogen 1/273 Helium 1/273 

Nitrogen 1/272.6 Carbon dioxide. / 1/270.4 

From this evidence, we can deduce that Boyle's and Charles's laws 
are rigidly applicable only over limited ranges of temperature, pressure, 
and volume. 

Ideal Gas. An ideal gas is one that, at constant temperature, in 
strict observance of Boyle's law, has the same P Rvalue regardless of 
variations in either the pressure or the volume. However, all known 
gases show variations in their behavior as compared to that of an ideal 
gas. For example, hydrogen and helium, at 0C., show a progressive 
increase in their PV values as either the pressure or the volume is 
increased or decreased. Oxygen and nitrogen, on the other hand, show 
a decrease in the value of PV with changes in either pressure or volume 
until a certain minimum is reached, at which time the pressure-volume 
product starts to rise. These deviations are slight and tend to disap- 
pear as the pressure of the gas approaches zero. It is common practice 
to study the behavior of actual gases in terms of their deviation from 
the behavior of ideal gases. 



60 QUALITATIVE ANALYSIS 

Nonideality of Gases and van der Waals' Equation. Two of the 
factors responsible for the nonideality of gases are (1) that molecules 
of a gas, contrary to the kinetic theory, do exert an attractive influence 
upon each other, especially at higher pressures, and (2) that whereas 
ideal gas molecules would merely be points in space, the molecules of 
an actual gas occupy a definite volume. 

It is clear that the attractive influence among molecules is negligible 
at low pressures because of the large amount of free space between 
them. At high pressures, however, the gas molecules approach one 
another, and the attractive influence is therefore more apparent. This 
force, whose magnitude varies inversely as the square of the total 
volume, tends to decrease the total volume of the gas. If this attrac- 
tive influence, at a volume of 1 liter, is symbolized by the letter a, 
then a/V 2 will express the attraction at any other volum. Since this 
attractive influence tends to decrease the volume occupied by the gas, 
it acts in conjunction with the externally applied pressure and must be 
added to this pressure. The total pressure of the gas is therefore 
(P + /7 2 ). 

At low pressures, the volume occupied by the molecules of a gas 
is negligible when compared with the total volume of the gas. There- 
fore, the free space and the total volume are essentially identical. For 
example, if the pressure exerted on a gas is doubled, the resulting vol- 
ume occupied by the gas would be one-half of the original volume. 
The volume occupied by the molecules is still negligible. The gas, 
then, obeys the simple Boyle's law. However, let us assume that the 
gas is under high pressure and that the volume occupied by the mole- 
cules is equal to J^o of the total volume. The volume of free space is 
then 20 1, or 19. If this gas is further compressed to one-half its 
present volume, the volume of free space is now 10 1, or 9, and the 
pressure is *% instead of 2, as would be expected from Boyle's law. 
This law must thqfefore be amended so that the volume expression 
takes into account the free space or true volume of the gas. The free 
space is equal to the volume of the gas minus the volume occupied by 
the molecules. This may be expressed mathematically as V fr, 
where b is the effective volume of the molecules. With this correction, 
Boyle's law now becomes P(V 6) = k. 

The simple gas law equation may now be written as (P + a/V 2 ) 
(7 6) = k. This is van der Waals' equation. In actual practice, it 
has been found that at ordinary pressures (1 to 2 atm.), the actual 
deviation of gases from Boyle's law is within the limits of experimental 
error. The widest deviation occurs for most gases in the ranges of 
50 to 100 atm. 



THE KINETIC THEORY OF MATTER 61 

It is apparent then that the simple form of Boyle's law, PV = fc, 
be used for ordinary work. The gases that deviate most widely, 
from the simple law are those that are most easily liquefied, since in 
these cases, the intermolecular attractive influence is strongest. 
Nitrogen, for example, which is liquefied with difficulty, departs less 
from Boyle's law than does carbon dioxide, which is liquefied much 
more readily. 

It should be noted that the two factors, attractive influence and 
molecular volume, which cause substances to deviate from the gas laws, 
act in opposite directions, the former decreasing the value of PV y 
the latter increasing the value of PV. There should be, therefore, 
for each gas a pressure at which these effects balance each other and the 
simpler law holds true. 

SOLIDS 

In sharp contrast to the general shapelessness of gases and the 
chaotic movements of the molecules comprising them, solids are char- 
acterized by a definite shape and form. The units comprising a solid 
substance are rigidly held together by forces of attraction that are 
considerably greater than those present in gases or liquids. As a 
result, solids possess considerable mechanical strength and rigidity. 
Two distinct classes of solid substances are known. The first class 
consists of substances that are crystalline in character, and the second 
class consists of the so-called amorphous or shapeless solids. 

Crystals. Crystalline material is characterized by a regularity 
of shape and a definite arrangement of the component particles. The 
crystals arise by growth, either from an evaporating solution or from 
a freezing liquid. During this process, there is a systematic growth 
from the unseen molecular particles to the point at which the outlines of 
the crystal are macroscopically evident. The visible crystal has the 
same form as at the very moment of genesis, and the completed crystal 
is merely an external manifestation of a specific and characteristic 
internal structure. We are familiar with the lacy, fernlike structure of 
ice or snow crystals on a windowpane. This is similar to viewing a 
very thin section of a crystal and is indicative of the general appearance 
of thin crystalline layers. Larger crystals, obtained from solutions, 
show definite three-dimensional outlines. 

Amorphous Substances. Among the amorphous substances, glass 
is a typical example. It does not exhibit any definite shape or form, 
and substances of this type are considered to be liquids in a very 
viscous state. There is, however, some ground for the belief that cer- 
tain amorphous substances are in reality microcrystalline powders 



62 QUALITATIVE ANALYSIS 

whose crystalline nature can only be determined by means of X-ray 
spectroscopy. The particles comprising amorphous substances are 
generally in a far less ordered arrangement than in crystals. They do, 
in fact, bear a closer resemblance to liquids than to crystalline solids. 

Crystal Structure. Since each crystal has a definite structure, the 
forces between contiguous particles are exerted in a definite direction. 
This definiteness of structure implies that each particle has a fixed 
position of equilibrium but does not preclude the conception that it 
is still free to move about in highly restricted areas and with a slight 
randomness of motion. We may assume that in the equilibrium state, 
the particles perform oscillations of very small magnitude about their 
fixed points of equilibrium. The application of heat will obviously 
increase the amplitude of these oscillations. At a certain temperature, 
these oscillations may become so great that the particles begin to collide 
with each other and may break loose from their fixed points of equilib- 
rium. When this temperature is reached, the solid structure disap- 
pears, and melting, or liquefaction, takes place. This temperature is 
the fusion or melting point. Since the particles comprising a solid 
exert considerable attractive force on each other, a certain amount of 
work must be done in separating the particles. The amount of work 
done, expressed as heat energy, is called the heat of fusion. This is 
defined as the number of calories required to convert completely 1 gram 
of a solid at its melting point to the liquid state, or vice versa. There 
are also other forces acting between particles in the solid state. These 
are the forces of repulsion that act to make solids only slightly com- 
pressible. The forces of repulsion increase much more rapidly than 
do the forces of attraction when pressure is applied to a solid, and 
therefore the distances between solid particles are only slightly reduced. 
Under ordinary conditions in a solid, the two sets of forces are in 
equilibrium. 

X Rays. Our primary interest in solids is not with the outward 
appearance of crystals but is rather centered on the internal arrange- 
ment of the particles making up the crystal. All crystalline substances 
may be fractured or split along definite planes of cleavage. The split- 
ting of sheets of mica into extremely thin plates is a good example of 
this phenomenon. This indicates/ of course, that each individual 
crystal has its component particles arranged in a very definite and 
characteristic structural pattern. The elucidation of the internal 
structure and arrangement of crystalline substances has been made 
possible by the use of X rays, which are light waves of extremely short 
wave length. The wave length of X rays is of the order of 10~ 8 cm., 
as contrasted with 10~ 5 cm. for visible light and 10 5 meters for radio 



THE KINETIC THEORY OF MATTER 



63 



waves. It has been shown that the distance between contiguous 
particles in crystalline substances is approximately of the same order 
of magnitude as the wave length of X rays; i.e., the particles are 
separated from each other by distances of the order of 10~~ 8 cm. 

The reasoning that led to these conclusions was based upon well- 
known facts obtained during experiments with so-called monochro- 
matic light. When light falls upon an object whose size is of the same 
order of magnitude as the wave length of the light, a perfect shadow is 



Source of 



X-Rctys 



SIH- 




Photographic __ 



FIG. 27. Diagrammatic sketch of the apparatus used to take X-ray pictures of crystals. 

not cast, but the light is diffracted or scattered. This phenomenon 
can be investigated by means of a diffraction grating. This grating 
consists of metal or some other material upon which a large number of 
very fine lines have been ruled. When monochromatic light is 
allowed to fall upon such a grating, the fine lines act as obstacles in 
the path of the light, causing it to be diffracted and giving rise to a 
number of images. 

. In a crystal, the orderly arrangement of the particles is the space 
or crystal lattice. Since the spacing of particles in crystals is approxi- 
mately 10~ 8 cm., the same order of magnitude as the wave length of 
X rays, the crystal may be used as a three-dimensional diffraction 
grating. Laue reasoned that the fundamental effect would be the 
same and proved it mathematically. If a thin beam of X rays was 
directed through a thin section of a crystal, the rays would emerge as 
a number of diffracted beams. If, now, the emitting rays were per- 
mitted to strike a photographic plate placed behind the crystal, the 
plate would exhibit a central spot because of the primary beam and a 
series of symmetrically spaced dots due to diffraction of the X rays 



64 QUALITATIVE ANALYSIS 

from the individual particles. Therefore, Laue concluded that the 
positions of the particles comprising the crystal could be deduced from 
a consideration of the position of the dots. 

These speculations were experimentally verified by Friedrich and 
Knipping a year later, with the use of the apparatus outlined in Fig. 27. 
These investigators obtained photographic plates wherein the struc- 
tural differences of the crystals were demonstrated by a wide variety of 
geometrical patterns, as illustrated in Fig. 28. Since the Laue dia- 




FIG. 28. Laue photograph of sodium chloride. 

grams were difficult to analyze, several later investigators introduced 
variations in the methods of obtaining the photographic plates with a 
consequent ease of analysis of the figures obtained. The Braggs sug- 
gested that instead of using the crystal as a transmission grating, it 
might be used as a reflection grating. In order to eliminate the neces- 
sity of using large, perfect crystals that were difficult to obtain, a 
method was developed independently by Hull, and Debyeand Scherrer, 
wherein a thin film of powder composed of fine crystalline particles was 
used in place of a single large crystal. 

The Space Geometry of Crystals. There are a large number of 
patterns that may be formed by the arrangement of solid particles in 
space. By far the most common arrangement found in both elements 
and compounds is the cubic structure, or lattice, which is the most 
symmetrical of all arrangements. In this simple geometrical form, 



THE KINETIC THEORY OF MATTER 



65 



there are three ways in which the particles may align themselves in 
space (see Fig. 29). 



Simple 
Cubic Lattice 





Face Centered 

Cubic Lattice 

FIG. 29. Simpler lattice types. 



Body Centered 
Cubic Lattice 



The first is the simple cubic lattice wherein the particles are placed 
at the eight corners of the cube. The second is the face-centered 
cubic lattice, differing from the first in that there arc also particles 
centered in each of the six faces. In the third, or cube-centered lattice, 
there is also a particle .in the 
center of the cube. Substances 
like KC1, AgCl, NH 4 I, and MgO 
are examples of the simple cubic 
type. Most of the metals are of 
the face-centered or cube-centered 
lattices. The face-centered is 
exemplified by Al, Cu, Ag, Au, 
and Pt. The last type is illus- 
trated by Na, K, Cr, Fe, and W. 

Structure of Sodium Chloride. 
In considering the use of X rays in 
the determination of crystal struc- 
ture, let us use as an example 
sodium chloride, which exhibits a 
cubic lattice. This substance, 
like rock salt, was, in fact, used by the Braggs in their studies. Theo- 
retically, the crystal lattice of this compound either could consist of 
sodium chloride molecules as the unit particles or have the components 
of the molecule (Na and Cl) separated. If the latter were true, the 
lattice would then consist of sodium atoms or ions combined with a 
lattice of chloride atoms or ions. From the evidence obtained with X-- 
ray methods, we can say that there are in sodium chloride two separate 
lattices. One of these contains only sodium ions (Na+) ; the other con- 
tains chloride ions (Cl~). Referring to Fig. 30, we see that each sodi- 




O Sodium Ions 
Chloride Ions 

FIG. 30. The crystal lattice (cubic) of the 
sodium chloride crystal. 



66 QUALITATIVE ANALYSIS 

um ion is surrounded by or shared by six chloride ions and that each 
chloride ion is surrounded by or shared by six sodium ions. The 
identity of the molecule in crystals of this type has disappeared, and 
each of the ions belongs equally to the six ions of opposite charge that 
surround it. The term molecule, then, must be applied to the whole 
crystal and not to a single unit of the crystal. In a great many cases, 
of course, certain crystals do have groups of atoms as the unit particles 
of their crystal lattices. In these cases, the groups of atoms are held 
together by primary or secondary valence forces. 

Isomorphism. In the late eighteenth century, the Abbe Haiiy, 
recognized generally as the father of crystallography, postulated that 
crystals that were identical in form must be identical in chemical 
composition and, conversely, that a difference in chemical composition 
implied a difference in crystal form. Shortly thereafter, many excep- 
tions were noted to this axiom. The work of Mitschcrlich, from 
1818 to 1820, contributed to disproving these postulates. While 
investigating the crystalline forms of various phosphates and arscnates, 
he noted that compounds of similar composition, containing the same 
numbef of atoms, had almost exactly the same crystalline form. In 
putting forward what is now known as the law of isomorphism, 
Mitscherlich stated his belief that the phenomenon was due to the 
ability of chemically similar but different compounds to exist in the 
same crystalline form. Originally, he considered that the same num- 
ber of atoms combined in the same way to produce the same crystalline 
structure regardless of the chemical nature of the atoms. In other 
words, the shape of the crystal was determined by the number of 
atoms and their manner of combination. As a result of later investiga- 
tions, Mitscherlich's statements were somewhat revised. The gen- 
eralization now accepted states that substances that have similar 
chemical composition very often have similar crystalline forms. Also, 
one atom of a crystal may be replaced by another atom without a 
change in crystalline form if the two elements are chemically similar. 
The alums are excellent examples of isomorphism. Common alum, 
KA1(SO 4 ) 2 -12H 2 0, and chrome alum, KCr(SO 4 ) 2 -12H 2 O, are definitely 
isomorphous. Other well-known examples are Epsom salts, MgSO 4 - 
7H 2 0, and zinc sulfate heptahydratc, ZnSO 4 -7H 2 0. There are, how- 
ever, a number of exceptions to this rule. The law of isomorphism 
does not give a true insight into the conditions necessary for isomor- 
phism. A number of salts have similar chemical composition but 
are not isomorphous (NaCl and KC1; KF and KBr). On the other 
hand, there are compounds like NaCl and AgCl that bear only a slight 
chemical resemblance to each other but that are isomorphous. Finally, 



THE KINETIC THEORY OF MATTER 67 



it has been found that isomorphism may occur even among substances 
that differ widely in chemical behavior (BaSCh and KMnO^. 

Obviously, then, complete identity of crystal structure is not a 
rigid criterion of isomorphism. One of the true criteria of isomorphism 
is the ability of substances to form mixed crystals. If two substances 
are truly isomorphous, they will both crystallize from solution together, 
the crystals always containing varying proportions of each substance. 
For example, CuS04 and FeS04 are isomorphous. When a solution 
containing these two substances is allowed to crystallize, the copper 
sulfate is always found to contain some ferrous sulfate, and regardless 
of the number of times the former is recrystallized, there is always 
some contaminant present. Fractional crystallization is obviously not 
sufficient to purify the- copper sulfate. However, if the iron is oxidized 
with nitric acid to the ferric condition, the crystals are no longer 
isomorphous, and they can be separated. The .sulfates of zinc and 
magnesium also defy separation by fractional crystallization. 

The study of the crystalline form and the properties of crystals is 
now made microscopically, this branch of analytical science being 
known as chemical microscopy. Those crystals whose physical 
properties are tho same in all directions are called isotropic, whereas 
those whose physical properties depend upon the angle in which the 
measurement is made are called aniso tropic. 

LIQUIDS 

Our discussion of gases and solids has provided us with an insight 
into the behavior arid structure of the two extremes in states of aggrega- 
tion. Liquids occupy a position intermediate between these two forms, 
and we might expect that they would resemble one or the other to some 
extent. As we shall see, liquids are fundamentally similar to gases both 
in behavior and structure. There must be, of course, some distin- 
guishing characteristics between the liquid and gaseous states. We 
have already seen that gases arc characterized by their ability to 
expand and fill the containers in which they are confined and that 
they possess, in addition, low density and viscosity. Liquids, on the 
other hand, have a definite volume for a given mass but have no definite 
shape. They have a greater density and viscosity than gases, and 
they possess a much lower degree of compressibility. In part, these 
differences may be explained by the fact that in gases the kinetic 
energy is much larger than the forces of attraction between the mole- 
cules, whereas in liquids the attractive influence is the greater. These 
variations, however, are not fundamental in character but are differ- 
ences in degree only. The chief reason for the demarcation between 



68 QUALITATIVE ANALYSIS 



the gaseous and liquid states of aggregation is probably due to the 
difference in the magnitude of the intermolecular forces of attraction 
and repulsion in the two states. These variations are governed chiefly 
by pressure and temperature relations, which may be so adjusted that 
the apparently striking differences are eliminated. 

We may assume that the unit particles of liquids are molecules (see 
Chap. VI). Obviously, the distance between contiguous particles is 
much less than in the gaseous state, and the free path between the 
molecules is very minute. Because of the closeness of the particles to 
each other, the forces of attraction exerted upon them are considerably 
greater than in the case of gases. This accounts for the slight com- 
pressibility of liquids. 

That liquids represent a highly condensed form of matter may be 
illustrated by the following example. When 1 ml. of water at its 
boiling point is converted to vapor at the same temperature, the volume 
occupied is about 1,700 ml. Since energy in the form of heat was 
required to bring about the change, we may safely assume that liquids 
possess^less energy than gases. The heat required to bring about this 
change is called the heat of vaporization. 

READING REFERENCES 

FINDLAY, A.: "The Spirit of Chemistry," Chaps. VI and VII, Longmans, Green 

and Company, New York, 1930. 
GETMAN, F. H., and F. DANIELS: " Outlines of Theoretical Chemistry," 6th ed., 

Chaps. II, III, and IV, John Wiley & Sons, Inc., New York, 1937. 
MACDOUGALL, F. H.: "Physical Chemistry," Chaps. Ill, IV, and V, The Mac- 

millan Company, New York, 1936. 
MELDRUM, W. B., and F. T. GUCKER: "Introduction to Theoretical Chemistry," 

Chap. II, American Book Company, New York, 1936. 

QUESTIONS 

1. If 275 ml. of hydrogen is collected over water at 27C. and 752 mm. pres- 
sure, what will be the volume of dry gas at standard conditions? 

2. If 150 ml. of oxygen is collected over water at 22C. and 765 mm. pressure, 
what will be the volume of dry gas if collected over mercury at 35C. and 740 mm. 
pressure? 

3. Compute the volume of SO 2 , collected at 25C. and 770 mm. barometric 
pressure, produced by the interaction of 5 g. of copper and concentrated H 2 S0 4 . 

4. In terms of the kinetic theory, distinguish the properties that serve to 
differentiate solids, liquids, and gases. 

5. A definite volume of nitrogen exerts a pressure of 0.25 atm. Three times 
as much oxygen is introduced into the vessel. What is the pressure of the mixed 
gas? If a volume of chlorine, equal to two and one-half times the original volume 
of nitrogen, is added to the mixture, what is the total pressure? 

6. A gas occupies a volume of 50 ml. at 16C. and 750 mm. pressure. At what 
temperature will the volume become 30 ml. if the pressure remains constant? 



THE KINETIC THEORY OF MATTER 



69 



7. In terms of the kinetic theory, account for each of the following: 

a. Crystalline form of substances 

b. Melting of crystalline substances 

c. Vapor pressure of a liquid 

d. Freezing point of a liquid 

8. Five hundred milliliters of a gas at S.T.P. was heatod until the volume 
increased to 750 ml. and the pressure became 775 mm. Calculate the temperature. 

9. Why do gases not obey Boyle's Law at high pressures? Why does Charles's 
law not hold rigidly at elevated temperatures? 

10. How many molecules are there in a liter vessel that has been evacuated to 
0.2 mm. at 0C.? 

11. Describe briefly the methods used to determine crystal structure. Draw 
the crystal lattice of KBr. 

12. Given that 1 mole of a gas occupies 22.4 1. at 0C. and 1 atm. pressure, 
calculate what the volume would be at 600C. and 300 atm. pressure. Use both 
the simple gas equation and van dcr Waals' equation. Calculate, also, the devia- 
tion of the values derived from the simple gas equation from those derived from 
van der Waals' equation. 

VAN DER WAALS' COEFFICIENTS 





a 


b 


Gas 


atm. X (liters) 2 


liters 




(moles) 2 


mole 


00 a 


3.592 


0.04267 


Cl> 


6.493 


0.05622 


He 


0.03412 


0.0237 


2 


1.360 


0.03183 



CHAPTER VI 

REACTIONS OF CHEMICAL COMPOUNDS 
I. PURE SUBSTANCES 

The most puzzling questions the student of chemistry asks himself 
are: Why do reactions occur? What is a chemical reaction or change? 
Why do substances sometimes react to form one compound, sometimes 
another, often a mixture of products and very often do not react at all? 
When is a substance a chemical compound, and when is it simply a 
mixture? These questions cannot be answered very simply. Some 
of them may be answered by definitions. These definitions are usually 
meaningless, however, if they are too narrow and futile if they are too 
broad. * Also, the answers to these questions vary with the year in 
which they are asked. One of the major tasks of theoretical chemistry 
is an attempt to formulate an intelligible answer to these questions. 

The earliest answers were anthropomorphic, i.e., an clement, A, 
"liked" another element, B, more than it "liked" a second, C; hence it 
combined with the first in preference to the second, C; or, to put it 
more scientifically, element A had greater affinity for element B than 
for C. This is not really an explanation but, rather, an observation; it 
does not tell us why the element A "likes" the element B more than C. 
In a sense, most of the explanations of reactions offered state the same 
thing. We are, however, gradually arriving at concepts that are 
providing a more fundamental answer to these questions. Whether 
we shall ever be able to answer this' question "why?" is problematical, 
since it eventually resolves itself into a question of philosophy. Never- 
theless, we shall attempt in this section to give as fundamental an 
answer as scientists have been able to wrest from nature. 

THE KINETIC APPROACH 

Let us consider two atoms moving about in a space of 1 cc. The 
likelihood of these two atoms colliding is remote, and since reaction is 
always preceded by collision, the likelihood of reaction occurring is 
more remote. But this is an improbable case, since the likelihood of 
such empty space occurring on the earth is extremely remote. It may 
occur in interstellar space, though not upon the earth. Let us, there- 
fore, consider a much more crowded space, such as occurs on earth 

70 



PURE SUBSTANCES 



71 



where atoms do collide and collide very frequently. In such a space, 
let us consider two molecules that are approaching each other at such 
velocities and in such a direction that a collision will occur. The 
question is, will these two molecules react? By a collision we do not 
mean that the nuclei actually collide. It is very doubtful if nuclei 
collide even in interatomic transformations, such as those caused by the 
cyclotron. What is meant is that the atoms or molecules approach 
sufficiently close to each other to enable the intermolecular forces to 
come into play. After the collision, the molecules either depart on 
separate paths or react, i.e., travel in a common path (sec Fig 31). 

Theoret/caJ cfrc/e 
within which mol- 
ecules act on each, 
other ^ 





Collision ana 1 exchange 

of en ray. No reaction. Compound 

formation 'I 1 

FIG. 31. Molecular collisions. 

The Governing Conditions. Now, to return to the question as to 
whether the two molecules will react. This question really does not 
mean much, since any two molecules or atoms will react, given the 
right conditions. The question, then, should really be, what conditions 
determine whether two atoms undergoing a collision will react? 
There are several factors, not necessarily independent, that determine 
whether reaction will take place. These are: 

1. How strong are the forces of attraction and repulsion between 
the atoms? 

2. What is the energy content of each molecule (e.g., velocity, 
among other things) ? 

3. How is the energy distributed in the molecule at the instant of 
collision; and as a corollary, what is the angle of approach between the 
molecules? 

To illustrate these points, we may select a very homely example. 
Assume that two boys are running down the street and wish to grasp 



72 QUALITATIVE ANALYSIS 

each other's hands. If they are approaching each other from opposite 
directions, their speed must be slow enough so that their momentum 
will not break their handclasp. If their speed is too great, they will 
clasp hands momentarily, but the momentum that their bodies have 
acquired will break the handclasp. The force of joining will, however, 
cause them to swerve. The greater their momentum the less will be 
the diversion caused by the momentary handclasp. If the boys are 
running in the same direction, one a little faster than the other, they 
could readily clasp hands as they came together. The handclasp would 
force one of the boys to run a little faster and the other to run a little 
more slowly, the exact deviations depending upon their respective 
weights (mass). If, on tho other hand, the boys were running in the 
same direction but at greatly different speeds and tried, as they passed 
each other, to clasp hands, the handclasp might not be strong enough; 
hence all that it would do would be to jerk the boys into another 
direction, slowing the fast one while speeding up the slow one. We 
think of atoms and molecules as behaving in a way similar to 
these Jboys. In a similar way the subject of thermal decomposition 
may be examined. If, when two atoms joined by a bond collide with 
another atom, the energy of collision is great enough, the molecule 
may break up, i.e., dissociate. 

Let us now examine the third factor, the distribution of energy in 
the molecule at the moment of collision and the angle of approach of the 
molecules. Returning to our example of the two boys, if each boy held 
in one of his hands a bag of books, it is evident that the boys could clasp 
hands only if, in addition to the conditions of speed, they approached 
each other in certain directions. This illustration applies to molecules 
when different parts of the molecules are more reactive than others. 
In considering the distribution of energy in the molecule, let us take 
the following example: If each of the two boys running down the street 
is juggling a ball, i.e., throwing it from hand to hand rhythmically, will 
they be able to clasp hands as they pass each other? This will, of 
course, depend upon which hand the ball is in at that moment, i.e., 
whether their ball juggling is in phase. In many molecules, the attrac- 
tion at a given point in the molecule may vary periodically because of 
the oscillation of the electrons or even because of the oscillation of a 
light atom. The probability, therefore, of a collision resulting in com- 
pound formation will depend upon whether the colliding molecules arc 
in phase. If one examines thcso requirements for reaction, one can se,e 
that, if a few billion collisions occur each second, it is highly probable 
that in $t least some of these collisions, the conditions will be just right, 
and reaction will occur. We may, therefore, state that if two sub- 



PURE SUBSTANCES 73 

stances are mixed, all possible reactions between them will occur 
to some extent. 

The Extent of a Reaction. The next question is, to what extent will 
each possible reaction occur? This is a very important question, since 
we know as a general rule that if we mix two particular substances 
under certain specific conditions, only very definite reactions will occur. 
The answer seems to be one of statistics; if we have several billion 
molecules having a particular energy distribution and undergoing a 
certain number of collisions each second, which reaction or reactions of 
these substances will predominate? 

First we must realize that, to judge by common criteria, quite a 
number of molecules must react before wo are aware of the fact. For 
instance, there are approximately 2.7 X 10 19 molecules per ml., i.e. 

27,000,000,000,000,000,000 molecule >s/ml. 

One of our most sensitive tests, the formation of lead sulfide from lead 
ion and sulfide ion, is barely visible if the concentration of lead ion is 
0.000025 per cent. This means that there must be at least 4.25 X 10 12 
(or 4,250,000,000,000 ions per ml.) ions of lead present before the reac- 
tion becomes visible. If the concentration is less, we shall not know 
whether the reaction has occurred. Hence a reaction may occur to 
such a slight extent that it will not be visible. 

THE ENERGY APPROACH 

The First Law. In addition to the statistical approach outlined 
above, the problem has also been approached from another angle. 
In this attack, there are several axioms. The first is the familiar 
statement that the energy in an isolated system is always constant, i.e., 
energy can be neither created nor destroyed. From this statement, 
knowing that energy can exist in many forms, we may say that in any 
system the 

Total energy = work energy + heat energy + chemical energy 

+ gravitational energy + electrical energy + 

In this system, one form of energy may be transformed into another 
form without changing the total energy. Now, each type of energy is 
composed of the product of two factors, the capacity factor and the 
intensity factor. Let us consider what this means. If we have twq 
blocks of wood, one of which has a mass of 1 Ib. and the other a mass of 
2 Ib., and if they are held 1 ft. above the ground, the amount of poten- 
tial energy that each block possesses, i.e., the amount of work that it 
would do if allowed to fall to earth, would, in the case of the body 



74 QUALITATIVE ANALYSIS 

weighing 1 lb., be 1 ft.-lb. and, in the case of the 2-lb. body, be 2 ft.-lb. 
The mass is the capacity factor, and the height through which it can 
fall is the intensity factor. In this example, the intensities are equal, 
but the energies are different, because the capacities are different. Let 
us take another example. We have a lake with a dam at one end, so 
that the water behind the dam is at a higher level than the water in 
the lake (see Fig. 32). Only the water behind the dam and above the 
level of the lake has the ability to do work; i.e., it possesses potential 
energy. If the dam is opened, the level of the water in the dam will fall 
to the level of the lake. The amount of water behind the dam and 
above the level of the lake is the capacity factor; the height is the 




FiG. 32. Potential energy stored behind a dam. 

intensity factor. If we examine this illustration, we see that the inten- 
sity of any given little portion of the water above lake level depends 
upon its height and, therefore, that the intensity varies with the height. 
When the dam has been opened and the total work accomplished, the 
capacity, i.e., the amount of water at lake level, has been increased and 
is equal to the total amount of water. The intensity, however, is now 
equal throughout, and this system can do no more work. 

The Second Law. This brings us to the second axiom or law, which 
states that all systems tend to reach equilibrium. If, in our last 
example, there were several dams surrounding the lake and the 
water stood at different levels behind each one, then, if we opened all 
the dams, the water level would become equal throughout. Thus we 
see that the system would have reached equilibrium. 

Types of Energy. Let us now consider what the components are of 
the various types of energy previously discussed. The first is work 
energy. The intensity factor of work energy is the pressure, and the 
capacity factor is the volume. From the discussion of gases in Chap. 
V, it may be seen that the amount of gas is the volume. In doing 



PURE SUBSTANCES 75 

work, this volume of gas must act against a certain pressure. Hence 
the intensity factor is the pressure and the capacity factor, the volume 
of the gas. In considering work energy in relation to solids and liquids, 
we define the intensity and capacity units in a different manner, more 
suitable for those states of aggregation. 

The intensity factor for thermal energy (the next form of energy) Ls 
the temperature, and the capacity factor is the entropy, the quantity of 
heat. The capacity factor for chemical energy is the mass, and the 
intensity factor is the chemical potential. These three types of energy 
are those that arc usually considered in chemical reactions. It is not 
necessary at this point to go into a more detailed development of these 
methods of analysis. Suffice it to say that by the use of such methods, 
it is possible to obtaip. a quantity called, variously, the free energy or 
the Gibbs free energy. If, in a given reaction, we subtract the free- 
energy content of the products from the free-energy content of the 
reactants, we obtain another quantity called the change in free energy. 
The algebraic sign of this quantity tells us whether the reaction can 
proceed, but it does not tell us if it will proceed at a measurable rate. 
The change in free energy is essentially the difference between the 
chemical energy of the products and the chemical energy of the react- 
ants. The extent to which any possible reaction will occur is then 
dependent upon the sign and the magnitude of this quantity. 

Correlation of Kinetic and Energy Aspects. This energy concept of 
reaction and the collision concept previously discussed actually supple- 
ment one another, as may be seen from the following example. The 
sign of the numerical value (which we can calculate) of the change in 
free energy tells us that, if we mix hydrogen gas with oxygen gas, they 
can react to form water, but it does not tell us how long it will take. 
We know that it will take a very long time, since mixtures of hydrogen 
and oxygen are quite stable ordinarily. This can be explained from a 
molecular viewpoint. In the hydrogen molecule, there are two atoms, 
as in the oxygen molecule, and it seems that before the hydrogen and 
oxygen can react, we must have single atoms. However, the kinetic 
energy of the molecules in the mixture, at ordinary temperatures, is 
such that the occurrence of single atoms is relatively rare, and conse- 
quently the likelihood of collision between individual atoms is very 
remote. Accordingly, the reaction will not take place at ordinary 
temperatures. If, however, we heat the mixture, the number of col- 
lisions and the energy of each collision increase, and therefore there are 
more single atoms and more collisions between single atoms. Since 
each such collision releases energy, the kinetic energy increases, and the 
whole material reacts rapidly. 



76 QUALITATIVE ANALYSIS 

THE CHEMICAL BOND 
PRIMARY VALENCE BONDS 

We have been discussing at great length the conditions necessary for 
reactions to occur, but we have said very little concerning the forces 
that hold the molecules together. We know that, when a reaction 
occurs, the two reacting atoms travel away on a common path, under- 
going a series of vibrations against each other, and in a previous chap- 
ter, we learned that several types of bonds exist between atoms. The 
question now is, what determines the type of bond that is formed, and 
how does the type of bonding influence the subsequent behavior of the 
molecule? 

This question is not one of the simplest to answer, for although the 
first result is that the molecules stick together (attract each other), the 
forces that cause them to attract each other will vary in strength, and 
this variation will be reflected in the subsequent behavior of the mole- 
cule. Let us consider what these forces are and how they affect the 
behavipr of the molecule. First we must remember that these forces 
are not independent but are only descriptions of extremes. Each of 
these types of linkage is represented in the "pure form" in some mole- 
cules, but other molecules fall into intermediate classes, where the 
forces acting cannot be classified so distinctly. 

The Homopolar Bond. As a first type of linkage, we shall consider 
the homopolar bond. Here the two atoms share a pair of electrons, 
one of which was contributed by each atom. The resulting molecule 
is more stable than either of the parts. Further, the two participating 
atoms attract the shared pair equally; i.e., the two electrons of the 
shared pair travel around both atoms and spend as much time under 
the influence of one atom as of the other. This describes one extreme 
case. 

The Heteropolar Bond. At the other extreme is the heteropolar 
bond, where one atom takes an electron away from another. There 
the transfered electron revolves solely around the host, and it does not 
enter the sphere of influence of the other atom at all. As a conse- 
quence, one atom is now positively charged, and the other is negative. 
The force that holds them together is the electrostatic force due to the 
negativity of one atom and the positivity of the other. This is the 
other extreme case. At one extreme, the atoms are very particular as 
to which atom they are attached to; at the other extreme, an atom 
desires only to be in the neighborhood of an oppositely charged 
atom, but it will readily trade its partner for another of the same 
charge. 



PURE SUBSTANCES 77 

The Semipolar Bond. One type of bond that is intermediate 
between these two extremes is the semipolar bond. In this bond, one 
atom that has a "lone pair" of electrons (i.e., two electrons that are not 
shared) will share this pair of electrons with another atom that- has 
space in its electronic superstructure for two more electrons. In this 
case, the pair of electrons travel in orbits around both atoms. If the 
atoms are neutral originally, that which donates the electrons becofnes 
a little more positive, and the one accepting the lone pair becomes a 
little more negative. This is logical, since the first atom, the donor of 
the lone pair, was neutral when it had all its electrons, but if it shares 
two of its electrons, it will no longer be sole " owner " of these negative 
charges and hence becomes a little more positive. Similarly, the 
acceptor of the lone pair becomes a little more negative. Of course, 
the molecule, as a whole, still remains neutral. If one of the participat- 
ing atoms has a charge, the same conditions exist in the molecule as 
before, but the molecule, as a whole, is electrically charged. 

The Valence Forces. The forces just described are usually called 
the primary valence forces and play the major part in the formation of 
the atomic aggregates that we call compounds. However, another 
series of forces exist that also cause atoms to travel in a common path. 
These forces are called the van der Waals' forces or, more usually, the 
secondary valence forces. They are, in general, much weaker than 
the primary valence forces and may be divided into three classes, which 
vary chiefly in their origin and in the strength of the resulting bond. 

SECONDARY VALENCE FORCES 

Rigid Dipoles. Let us assume that we have a neutral molecule 
having a bond that is intermediate between a heteropolar and a homo- 
polar bond. Because of the effects previously described, the molecule, 



c- 




FIG. 33. Dipolo compounds: two types. 

although neutral as a whole, will have one part that is more electro- 
positive and another part that is more electronegative than the adjoin- 
ing sections of the molecule. As a consequence of this fractional 
charge, if two of these molecules approach one another, they will 
align themselves in such a way that the negative charge of one molecule 
is as close to the positive charge of the other molecule (and vice versa) 
as the geometry of the molecule will permit. Thereafter the two 



78 QUALITATIVE ANALYSIS 

molecules will travel together until sundered by a collision. Com- 
pounds possessing bonds of this type are called dipole compounds. 

Induced Dipoles. Another variation of this force exists when a 
compound possessing a dipole approaches a molecule having no dipole, 
in which the outer electrons, however, are quite mobile. If the nega- 
tively charged portion of the dipole molecule approaches the neutral 
compound, it will repel the mobile electrons and convert the neutral 
molecule into a temporary dipole. If the positively charged portion 
approaches, it will attract the mobile electrons and also form a tem- 
porary dipole; i.e., the dipole molecule will induce a charge or dipole in 
the neutral molecule. The induced dipole and the permanent dipole 
will now attract each other and so form another type of dipole com- 
pound. This type of force is, of course, weaker than that between two 
permanent dipoles. 

Dispersion Forces. The third type of van der Waals' force is the weakest 
type and was postulated to explain the fact that rare gases, like argon and neon, 
at high pressures form liquids and solids. For this phenomenon to occur, some 
type of attraction must exist between the atoms. The primary valence forces arc 
ruled out, since the rare gases form no stable compounds. We know, also, that 
these molecules are symmetrical; therefore, no dipoles or induced dipoles can exist; 
but forces do exist! To explain this, London postulated that when the atoms are 
very close together, they tend to arrange themselves so that their electronic orbits 
arc in resonance with each other. This third variety of van der Waals' forces is 
called the dispersion force. This postulation has been subjected to experimental 
verification and has been found valid. 

As a last word to this section, let the reader beware lest he think of 
these types of bonds as separate entities, remote and distinct. After 
all, all those bonds are the result of the interaction of two electrical 
systems, each of which is composed of positive charges and negative- 
charge clouds of variable density electrons, if you choose. The result 
is a net attraction or repulsion. From the magnitude of this effect 
and from the behavior of the resulting system we then reason back- 
ward and classify the types of linkages. 

REACTIONS BETWEEN GASES 

We have now discussed in a general way why reactions occur, but 
for many purposes we wish to know not only why they occur but also 
to what extent they occur. This question can be dealt with, for gases, 
in a statistical manner. 

Equilibrium Constants in Terms of Concentration. Let us suppose 
that we have two gases, A and B, in a vessel. We know that they will 
react to form the compound AB. We know, also, that the compound 
AB will break up again into A and B. We know, too, that after some 



PURE SUBSTANCES 79 

time, the reaction will come to equilibrium; i.e., AB will be formed as 
fast as it decomposes. When this equilibrium has been established, 
how much of A, B, and AB will be present in the vessel? The reaction 
may be represented as 

A + B 



The equation is written with a double arrow to indicate that there is an 
equilibrium. The concentration, i.e., the number of gram moles per 
liter of A may be represented by C A ; the number of gram moles per 
liter of B, by C B , and the number of gram moles per liter of AB by 
C AB . The speed with which A will combine with B is given by the 
formula 



B = ki(C A X C B ) 

That is, the speed of this reaction will be determined by a constant fci, 
which takes into account the various factors outlined previously (such 
as the velocity and energy content of the molecules, the energy distribu- 
tion in the molecules, and the number of collisions per unit of time) 
multiplied by the concentration of iho two ivactants. Similarly, the 
speed of the other reaction, the decomposition of AB will be given by 
the formula 

Spccd Ab = ^(CAB) 
At equilibrium 

Speed A +B = specdAB 
therefore 

fci(C A X C B ) = /c 2 C AB 
or i 

_ C A B _ _ /Cj 

C A X CB &2 

Since & 2 and fci are both constants, wo may substitute another constant 
for them 



A/* 2 C A X CB 

K is the equilibrium constant for the reaction 

A + B ^ AB 

One might ask why is the constant taken as being k\/kt rather than kz/k\. 
This is purely conventional; both are correct, but since confusion might arise, K 



80 QUALITATIVE ANALYSIS 

is so taken that in the reaction A + B ^ AB, the concentration of the substances 
on the right side of the equation appear as the numerator of the term 



Cleft side 

For the reaction AB ^ A + B, the decomposition of AB, the equation is 



CAB K 

Example 1. Let us take the decomposition, in the vapor phase, of 
phosphorus pentachloride. The reaction is 

pcis ^ PCU -f- cu 

The equilibrium constant is 

-rr ^PCl3 X v>Cl2 

/V c -T y 

t'PCU 

The subscript c is added to K to indicate that the constant is con- 
cerned with molar concentrations. Now we shall calculate K c for the 
foregoftig reaction. The facts, as obtained by analysis, are that at 
250C. and 1 atm. pressure, the phosphorus pentachloride is 80 per cent 
dissociated. Therefore, if we started with 1 gram mole of PC1 5 , at 
equilibrium (250C.), we should have 0.8 gram moles of PC1 3 , 0.8 gram 
moles of C1 2 , and 0.2 gram moles of PC1 6 . Now we must calculate the 
concentration in gram moles per liter. One mole of gas would occupy 
22.4 liters at 0C. or 273 A. and 1 atm. pressure; at 250C. or 523A., it 
would occupy 

523 
22.4 X HTQ = 42.9 L, volume of 1 mole at 250C. and 1 atm. pressure 



At equilibrium, we have for every mole with which we started 
0.2 mole PC1 5 , 0.8 mole C1 2 , and 0.8 mole PCU, or 0.2 + 0.8 + 0.8 = 
1.8 moles. Therefore, the total volume at equilibrium would be 

42.9 X 1.8 = 77.2 1. 
The concentration of C1 2 is 

Cci t = ~s~o = 0.0104 mole/liter 
77. 6 

and 

0.0104 mole/liter 



7i . 



= 0.0026 mole/liter 



PURE SUBSTANCES 81 

therefore 

K Crci a X Cci 2 (0.0104) (0.0104) 
c Cpci 5 0.0026 

Example 2. Let us take another typical example, the decomposi- 
tion of hydrogen iodide gas. Hydrogen iodide at 448C. decomposes 
partially into hydrogen and iodine according to the equation 

2HI ^ II 2 + I 2 
The equilibrium constant of this reaction is 

rr __ CH Z XClg 

A '~ (Cm) 2 

If we start with 2 moles of hydrogen iodide, the volume of the 2 moles 
at 448C. will be 

721 
2 X 22.4 X ~ = 118.3 1. 



Since the number of moles is not changed by the reaction, the volume of the 
mixture will be constant. 

After equilibrium has been established, the gram moles of HI remaining 
will be 

2 - 2x = moles of HI 

and since the gram moles of H-2 equals the gram moles of lz and since 
this is unknown 

x = g. moles of H 2 
x = g. moles of I 2 

The concentration of the various constituents is now 

2 - 2x 

HI " 118.3 
~ x 



" l2 118.3 

r x 
CHZ "" 118.3 

and 

x x 



7 H2 X Ci 2 118.3 118.3 x 2 



(Cm) 2 - /2-2xV (2-2*)* 



(2 -a*y 

\ 118.3 / 
At 448C., hydrogen iodide dissociates to the extent of 21.99 per cent; 



82 QUALITATIVE ANALYSIS 

since x is the amount of dissociation, substituting 0.2199 for x gives 

K- (0.2199) 2 

* ~ [2 - 2(0.2199)f 

Example 3. This problem may also be reversed. Let us place in 
the reaction flask a certain volume of iodine, a, and a certain volume of 
hydrogen, 6; raise the temperature to 448C., and allow the materials to 
come to equilibrium. At equilibrium, a certain amount, z, of iodine 
will have reacted with the same amount, x, of hydrogen. Therefore, 
the concentration of hydrogen present will be b x, and that of iodine 
will be a x. The reaction of the x moles of iodine with the x moles 
of hydrogen will give rise to 2x moles of hydrogen iodide. Therefore, 
for the reaction I 2 + H 2 ^ 2HI, we could write 



K' = 



Cni 2 



i z X Cn 2 



but K'o, = jr- where K c is the equilibrium constant for this equation 

written backward. Hence 

JT Ci 2 X Cn 2 

Ac - jfY-i 

CHI 

And since 

Ci, = a x 

CH, = b x 

CHI = 2x 

we could substitute these in the equation 

7T (a s)(6 x) _ ab bx aa; + x 2 



and solving for x 

4K c x 2 = ab ~bx - ax + x 2 
and 

z 2 (l - 4K C ) - x(a + 6) + ab = 

Applying the quadratic formula, we get 



= 



2(1 - 



If the original concentrations were 2 moles of hydrogen to every mole of 
iodine, a = 1 and 6 = 2, and since we know that 

K c = 0.01984 



PURE SUBSTANCES 83 

we can solve the problem 



(1 + 2) + V(l + 2) 2 - 4j(l X 2)[1 - 4(Q.01984)]| 
x 2[1 - 4(0.01984)] 

= 3 \/9^ =r 8"+ 0.635 = 3 \/17635 3 1.28 ' 
X 2 - OT.159 1.841 1.841 

x = 2.32 or 0.934 

The concentration of HI is 

CHI = 2z 

CHI = 1.868 moles 

The first answer, 4.64, is manifestly extraneous and is impossible, since 
the maximum possible yield if the reaction proceeded to completion 
would be 2 rnoles of HI, since only 1 mole of iodine was used and 1 mole 
of iodine can yield no more than 2 moles of HI. Therefore 

Cm = 1.868 moles 

C H2 = b - x = 2 - 0.93 i = 1.066 moles 

Ci 2 = a - x = 1 - 0.934 = 0.066 mole 

The last problem may raise the question as to why we both square 
and double the concentration of the HI. Kxactly what is meant by 
this statement may not be apparent; hence let us choose an example. 
If we had the following reaction 

A + B^D + E 
the equilibrium constant would be 

CD X CB 



K c = 



CA X CB 



and if we started with 1 mole of A and 1 mole of B, the concentration at 
equilibrium would be: 

CD = x C E = x 
C A = 1 - a?, C B = 1 - x 
and 

(*)(*) _ x* 



" c (1 - x)(l - x) (1 - xY 

If now the molecules of D and E were the same, i.e., D and E are 
identical, the equation would be 

A + B ^ 2D 



84 QUALITATIVE ANALYSIS 

and 



rt 



C D 2 



C A X CB 

Again, if we started with 1 mole of A and 1 mole of B, the concentra- 
tions at equilibrium would be 

C A = 1 ~ x, C B = 1 - x 
CD = 2z 

Substituting, we should get 

, (2ffi 4ar> 

c (1 - x)(l - x) (1 - z) 2 

By solving these two equations and equating, we get 

K-c == TV 

# = 



(T - xY 

4x* 



(1- 



4^ (1 - xY 
~ = K 

T7~t A rr 

K c = 4A C 

We see that if the two molecules are alike, the equilibrium constant 
is four times as great as the equilibrium constant when the two mole- 
cules are different. The reason for this will be apparent from the fol- 
lowing discussion. 

Let us consider four random collisions of the product molecules. 
When the two molecules are different the collisions would be 

D and D D and E 
E and E E and D 

In this case, only the two collisions occurring between D and E will go 
to form A and B. Let us now consider the other case, where D and E 
n,re the same (only D molecules are in the product). The four 
random collisions would be: 

D and D D and D 
D and D D and D 

That is, in the second case, there will be twice as many collisions leading 
to reaction as in the first; or, to put it differently, if the molecules are 
alike, the effective concentration is twice as great. Therefore, we can 
write the equilibrium constant of this reaction as 



PURE SUBSTANCES 85 

r r 

jy-f ^D ^D 



and since the effective concentration of C D is 2x 



_ _ 

~ (1 - x)(l - x) - (1 - xY 

Equilibrium Constants in Terms of Pressure. In most gases, the 
simple gas law holds 

PV = nRT or P = y RT 

Now n/F (the number of moles per unit volume) is the definition of 
concentration; hence, we may write 

P = CRT or 

Therefore, in any equation involving gas concentrations, P/RT can bo 
substituted for C. Let us see what would happen if we did this in our 

first example: 

PC1 5 ^ PC1 3 + C1 2 

V V/l'CJIs X CcljJ 

A c = -- ^ - 

CPCU 
Now 



_ 

PCI. - T 



where ppci 8 is the partial pressure of PC1 3 , etc. 



RT 
or 

1 

?T P pcla X Pci 2 v 



_ 

RT 

_ PPCI X 





86 QUALITATIVE ANALYSIS 

We can readily see from this relationship that 

I, X Pch 



is a relationship quite analogous to 

Cpcia X Celt 
Cpcii 

Therefore, it is quite logical to call this K p or 

Kp = P^^P^ = Kc(RT ) 



There is always a relationship between K c and K p , but this relationship 
is determined by the reaction and must be derived for each case. The 
exact form of the relationship can be obtained by substituting 

p = CRT or C = 



in the equilibrium equation. 

Example 4. Let us now do a problem involving K p . 

PC1 5 ^ PC1 3 + C1 2 

v _ PPCI, X PCM 
j\'p ~ 

PPCI, 

Tf we start with 1 mole of PC1 5 and x moles dissociate, at equilibrium we 
shall have 1 x moles of PC1 5 and x moles of PC1 3 and x moles of C1 2 . 
The total number of moles present will be 

(1 - *) + (x) + (x) = l+x 

Each gas in this mixture exerts its own partial pressure independently 
of the other gases, and the exact magnitude of this pressure depends 
upon the amount of the gas present. Since we know the amount of 
each gas present and the total pressure, we can calculate the fraction 
of the total volume occupied by each gas and so calculate the partial 
pressure. 

1 - x 



Mole fraction of PC1 5 = 
Mole fraction of PCU = 



l+x 
x 



Mole fraction of CU = i r- 



PURE SUBSTANCES 87 

The partial pressure of each gas is the mole fraction of that gas multi- 
plied by the total pressure, where P = total pressure 



Substituting these values in the equation, we get 

x p v ' X P ? P2 



x 



2 



l+x 
A,, = 



1 - x p l_-_x 

l+x 1 + x J 

p 



(1 - x) (1 + ar)(l - x) 

We know that PC1 5 at 250C. and 1 atm. pressure is dissociated to the 
extent of 80 per cent. This means that x, the amount dissociated, is 
0.8 and that P is 1 atm. Substituting these values, we get 

K = _ W_ _ m 
p - -F0.8) 



K = - - _ 

p ~ 0.36 ~ 



From this value, let us calculate K c . As we have seen on page 86, for 
this reaction 

Kc = Kp X Rf 

T = 273 + 250 = 523A. 

R = 0.0821 l.-atm. 

1 1 



K c = 

K c = 0.0414 
Example 5. Let us calculate K p for the reaction 

2HI - H 2 + I 2 
For this reaction 

K = P H X PI, 

P ~~ PHI 2 



88 QUALITATIVE ANALYSIS 

and from the gas laws 

p = CRT 
and 

* x gr > CH * x 



therefore 

K p = If c 

Since we have previously calculated X r , we need do no more. 

REACTIONS BETWEEN SOLIDS 

This heading usually prompts the questions: Do solids react with 
each other? Is not the reaction usually preceded by a change to the 
liquid state? These questions may best be answered by citing a simple 
experiment that can be verified with little trouble. If one takes solid 
potassium iodide and mixes it with solid mercuric nitrate and grinds 
them together in a mortar, one will observe that the mixture that was 
originally white becomes orange-red. In fact, it is not even necessary 
to grind them together; merely put them into a bottle, shake vigor- 
ously, and the mixture will change its color. The reaction that has 
taken place is the formation of mercuric iodide: 

2KI + Hg(N0 3 ) 2 -> HgI 2 + 2KNO 3 

Now that the fact that solid reactions take place has been verified, how 
can we explain them, knowing that the ions in these solids are relatively 
fixed in the crystal, i.e., that they are vibrating around fixed points in 
the lattice? The explanation is not too difficult. It depends upon the 
bond energies, i.e., the magnitude of the attractive force between the 
ions in the crystals. 

When we rub two solids against each other, the ions at the surface 
of one crystal come into fairly close contact with the ions at the surface 
of the other crystal. If two of the ions in such a "collision" attract 
each other with a greater force than they do their original partners, 
they may break away from their old partnership and form new com- 
pounds. However, the process occurs only at the surface of the crys- 
tals, and soon the process will end because of the fact that the surfaces 
become completely used up. If we grind the compound, we break 
up the crystals, forming new surfaces that can react, and we also bring 
these new surfaces into more intimate contact with each other and 
thus form more of the new compound. Therefore, grinding potassium 
iodide and mercuric nitrate gives the mixture in the previous experi- 
ment a redder color than merely shaking them together. 



PURE SUBSTANCES 89 

Adhesion. Many solids, when brought together, stick or adhere. 
The reason for this phenomenon is comprehensible if we consider that a 
solid reaction takes place at the points of contact between the two 
surfaces forming a compound. We do not usually consider this 
phenomenon as a type of compound formation, but adhesion is never- 
theless a variety of compound formation that occurs at surfaces, and it 
is caused by the same forces that form compounds. Usually 'these 
forces are the secondary valence forces, or van der Waals' forces. 

REACTIONS BETWEEN LIQUIDS 

Pure liquids are not encountered too frequently in inorganic 
chemistry; especially is this true of elementary inorganic chemistry 
and work at normal temperatures. The notable exception is, of course, 
water. Liquids may be defined as substances in which the forces of 
thermal agitation have overcome the forces that maintain the crystal 
shape. Although permanent stable structures do not exist in liquids, 
temporary structures of short life do (exist. The forces existing 
between the molecules are still large (compared to gases), and hence the 
molecules are close enough to each other so that their free paths are 
extremely small. In liquids, the molecules vibrate about a fixed 
position for a short while and then move to another position, where they 
again oscillate about a fixed point. It can thus be seen that liquids 
resemble both solids and gases. This similarity is further apparent 
in the reactions of liquids. 

Before proceeding to the reactions of liquids, we should discuss the 
forces that hold liquids together. As is usual in chemical phenomena, 
the forces cannot be divided into classes as wfc divide sheep and goats, 
but we can describe the extremes of behavior. One extreme type of 
behavior is that exemplified by liquids in which the binding forces are 
electrostatic; i.e., each particle, now called ion, has an electrical charge, 
either positive or negative. In such liquids, each ion is surrounded by 
an atmosphere of oppositely charged ions. However, none of these 
oppositely charged ions is firmly fixed in position, and none is close 
enough, except momentarily, to form a heteropolar bond. The atmos- 
phere of oppositely charged ions that surrounds each ion is not fixed but 
varies at random. If we now add another liquid of the same type, 
several things may happen. First, if the liquids are added one to 
another slowly, they will form two layers and then slowly diffuse until 
they are thoroughly mixed. This diffusion in liquids is relatively 
slow compared to diffusion in gases and depends upon the viscosity 
of the liquid. The viscosity of a liquid or, in simpler language, the 
ease with which a liquid will flow, is only a measure of the forces of 



90 QUALITATIVE ANALYSIS 

attraction between the molecules. To demonstrate the slowness of 
diffusion in liquids, one can perform the following experiment. If one 
places some solid copper sulfate in the bottom of a cylinder and care- 
fully fills the cylinder with water, the water will be colored blue about 
the crystals quite rapidly, but the blue color will progress up the cylin- 
der very slowly, requiring a period of several weeks before the whole 
cylinder of liquid is colored. 

B- B- A f A+ B- B- C + A+ A+ 

B-A+B- A +B- A + D ~ A+ D- A + B ~C+ 

B - # ff B" D~ " C+ 

B~ r +A+ 

B" B" C+ 

One Liquid ~ 

Mixed Liquids 

FIG. 34. Atmospheres surrounding ions. 

Let us now consider what happens when two liquids are mixed. 
Again we must deal with extremes; at one extreme nothing happens 
except that the atmospheres arc now more heterogeneous than they 
previously were. The equation for this reaction can be written as 

(A+ + B-) + (C+ + D-) -> A+ + B" + C+ + D- 
The other extreme can be symbolized in an equation as 

(A+ + B-) +. (C+ + D-) -* A+D- + C+B- 
or 

(A+ + B-) + (C+ + D") -> A+D- + C+ + D- 

In this case, let us examine an ion A + , surrounded by an atmosphere 
of B~ ions. Along comes an ion D~~, which moves into the atmosphere 
and, after an instant, grabs ion A + and holds it very tightly, at the same 
time repelling all the other negative ions that come into the vicinity. 
This is, of course, compound formation. If this happened every time 
an A+ ion approached a D" ion, the reaction would go to 100 per cent 
completion. Since we know that this docs not occur, the description is 
justly labeled the extreme case. Actual substances fall between one 
extreme, where only heterogeneous atmospheres result, and the other 
extreme, where reactions go to completion. 

We have thus far discussed liquids held together by electrostatic 
forces. At the other extreme are liquids held together by van der 
Waals* forces. In these liquids, the particles (molecules) are neutral, 



PURE SUBSTANCES 91 

and their arrangements are more complex than in purely ionic liquids. 
Reactions between two liquids of this type will vary according to the 
strength of the van der Waals' forces and the strength of the bonds in 
the molecule. However, since such liquids are only rarely encountered 
in qualitative analysis, we shall not discuss reactions between them. 
Suffice it to t say that most real liquids fall somewhere between the two 
extremes,* and hence most reactions between pure liquids are not too 
simple. 

READING REFERENCES 

GETMAN, FREDERICK N., and F. DANIELS: "Outlines of Theoretical Chemistry," 

6th ed., John Wiley & Sons, Inc., New York, 1937. 
LATIMER, W. M., and JOEL H. HILDEBKAND: "Reference Book of Inorganic 

Chemistry," rev. cd., The Macmillan Company, Now York, 1940. 
MARK, H.: "Physical Chemistry of High Polymeric Substances," Intcrscicnce 

Publishers, New York, 1940. 

QUESTIONS 

1. At 49.7C. and 497.75 mm. pressure, N 2 O 4 decomposes into NO 2 to the 
extent of 49.3%. Calculate K c and Kj,. 

2. At 3000C. and 1 atm. pressure, 40% of the molecules of CO 2 are decomposed 
into O2 and CO. Calculate K r and K p . 

3. The equilibrium constant of 2HI ^ I 2 + IT 2 is equal to 0.01984 at 448C. 
If we start with 4 moles of hydrogen and 1 mole of iodine and permit the system 
to attain equilibrium, compute the concentrations of the components of the 
equilibrium mixture. 

4. For the reaction CaC0 3 ^ CaO + C0 2 , what would be the effect on the 
equilibrium of an increase in pressure? 

6. For the reaction 2N11 3 ^:N 2 -f 3H 2 , what would be the effect on the 
equilibrium of an increase in temperature? 
6. Explain why paint sticks to a wall. 



CHAPTER VII 

REACTIONS OF CHEMICAL COMPOUNDS * 
II. REACTIONS WITH SOLVENTS 

Although the reactions between pure gases, pure liquids, and pure 
solids are very interesting, they are not encountered too frequently in 
elementary inorganic chemistry. Most of the common reactions take 
place in aqueous solutions. However, before one can understand what 
happens during reaction in aqueous solutions, one must understand 
what happens when substances dissolve. 

Solutions have been studied for a great many years, but the exact 
elucidation of the facts about solutions are still far from complete. 
However, we do understand to a certain extent, at least dilute 
solutions. We shall try to present here the modern views regarding 
the architecture if it may be called that of solutions. 

Why Substances Dissolve. We are sure that every chemist and 
student of chemistry has asked himself at some time or another, why, if 
I place substance A in water, will it dissolve, whereas if I place sub- 
stance B in water, it will remain there unchanged? Of course, this 
question could be and was, in fact, answered simply for a long time in 
the following way. A substance dissolves because of its innate nature; 
or, to put it into other words, a substance dissolves because it is soluble. 
However, we know, and the chemists who gave this answer knew, that 
this was not an answer but only a dogmatic statement. Nevertheless, 
so little was known at that time that it was not profitable to pursue this 
question too far, since with only a scant handful of facts at their 
command, the chemists' answers soon degenerated into metaphysics 
and mysticism. Enough facts have been gathered at present, however, 
so that we can present a fairly intelligible picture of what happens 
when a substance is placed in a solvent. 

Let us now take a concrete example. If we place a lump of copper 
sulfate in a beaker of water, we note that it dissolves gradually, turning 
the whole solution blue. We can ask, what has happened? Why does 
the copper sulfate dissolve? In the crystal of copper sulfate, we have a 
framework composed of tetraquocupric ions, [Cu(H 2 O) 4 ] ++ , and sulfate 
ions, SO 4 ", connected by a water bridge, thus giving the empirical 
formula CuSO4-5H 2 O. A schematic drawing of the structure is shown 

92 



REACTIONS WITH SOLVENTS 



93 



in Fig. 35. (Of course, this picture does not show the three-dimensional 
structure of copper sulfate or the other bonds that help it keep its 
three-dimensional form.) 

The water molecules in the liquid water hit the molecules at the 
surface of the crystal, and, although many of them rebound, some of 
them stick and form a surface compound or layer. We know that this 
happens in most cases when a solid is in contact with a gas or a liquid. 
The reason for this adhesion of a layer of material is fairly easily 
explained. In the sulfate ion, the oxygen atoms are relatively negative 
as compared to the sulfur. In Fig. 36, these fractional charges are 



//// 

i o \ 



H 



0\Cu0 \0\S\0\ 



H* *H 

FIG. 35. Schematic picture showing the 
water bridge between the two ions in 
CuSO 4 -5H 2 0. 









I 


:o 


s 


o: 


L i 







i 



FIG. 36. Dipoles within the 
SOV" ion. 



represented by arrows, the hoad being the negative and the tail being 
the positive end. Of course, the ion as a whole has a negative charge, 
and it is not a dipolc, because the individual di polos of the bonds are 
symmetrical and so cancel each other. However, each oxygen can 
attach itself, under propitious circumstances, to other dipolos, either 
permanent or induced. In the interior of the crystal, all these forces 
balance because of attached groups on all sides. However, at the sur- 
face, some of these forces are available, since certain sides of the ions are 
exposed and so can attach themselves to any available molecules or 
ions. Of course, Fig. 37 is only a two-dimensional picture and repre- 
sents a very simple case, but it serves to show that at the surface there 
are forces available, both fractionally positive and negative. These 
forces attract a monomolecular film to the crystal. That film may, in 
turn, attract another film and so on, but the extra films do not usually 
stick so well as the first. This is the first step, and it happens whether 
the substance is soluble or not, just as long as the water will wet the 
solid. If the water does not wet the solid, then for some reason, the 
disposition of the forces at the surface is such that the solid is more or 
less indifferent to the water. 

Since the molecules in the crystal are in constant vibration, at some 
time or another, some of the forces that were neutralized originally 



94 QUALITATIVE ANALYSIS 

in the body of the crystal attach themselves to solvent molecules; i.e., 
molecules of water work their way between the ions and insulate them 
from each other, thus breaking the bonds holding them together. 
Finally, as this process continues, all the bonds are broken, and whole 
ions break away and, being surrounded by atmospheres of solvent 
molecules, are buffeted about and so diffused throughout the liquid. 

Crystallization. The reverse process also takes place. If enough 
of the dissolved ions are present in a solution, they will collide at inter- 
vals, and if the collision has the right energy and if the geometric 



_ _ 

/ t 

\ /_ \ / L_\ / 
/\ /\~* /\ 



Surface 



NX 

~ /\ 



FIG. 37. Schoraatio picjturo of internal and surface dipolcs in a crystal. 

conditions are right, they may squeeze out of the way some of the 
atmosphere surrounding each and so permit their forces to act upon one 
another. This bond may break during the next collision, but if suffi- 
cient ions are present in solution and their collisions arc frequent 
enough, groups of ions will form. Those groups of ions may grow in 
size, and if they reach a size of approximately 10~ 7 cm., they begin to 
appear as crystals. It is thus apparent that crystals have such definite 
shapes because they reflect the molecular distribution of forces. 

Solubility. When these two tendencies of the solid to dissolve and 
the ions to crystallize are equal, the solution is saturated. By adopt- 
ing appropriate units, we may arrive at a figure that we call the solubil- 
ity of the solid. This is usually expressed as the number of grams that 
can dissolve in a liter of water or the number of moles that can dissolve 
in a liter of solution ; but other modes of expression may be used. How- 
ever, solubility is merely a convenient method of measuring this 
equilibrium in the solution. 



REACTIONS WITH SOLVENTS 95 

SOLUBILITY CALCULATIONS 

Percentage Solutions : 

Example 1. How would you prepare a 50 per cent solution of 
AgNO 3 ? Dissolve 50 grams of AgNO 3 in 50 grams of water. 

Example 2. How would you prepare a 3 per eent solution of 
cadmium acetate? Dissolve 3 grams of cadmium acetate in 97 grams 
of water. 

Molar Solution. A 1M solution contains 1 gram mole of solute per 
liter of solution. 

Example 3. How would you prepare a 1 M solution of ZnSCVOHoO? 
The molecular weight of ZnS(VOH 2 <) = 209.53. Dissolve 209.53 
grams of ZnSOrOIW) in water, and make up to 1 liter of solution. 

Example, 4. How many grams of BaClo^II .() are present, in 27 nil. 
of a 1.34M solution? Molecular weight of Ba(V2II 2 (> = 214.31. 
One liter will contain 1.34 X 244.31 grams, and 27 ml. will contain 

27 

X 1.34 X 244.31 grams = 8.81 grams. 



Molal Solution. A 1 molal solution contains I mole of solute per 
1,000 grams of solvent. 

Example 5. How would } r ou prepare a 1 .05 molal solution of KOI? 
Molecular weight of KC1 = 71.50. Dissolve 1.05 X 74.50 = 123.02 
grams of KOI in 1,000 grams of II 2 O. 

Example 0. What weight of LiF 1 is present in 110 grams of a 1.2 
molal solution? Molecular weight of LiF = 25.91. 1.2X25.94 
grams in 1,000 grams of I U>() = 1,031.13 grains. 31.13 grains of 
LLK =c= 1,031.13 grams of solution, 1 x grams of LiF =0= 110 grams of 
solution. 1 

3L13 = 1,031.13 
~ x 110 " 

31.13X110 000 <_. 
*-- 1,031. 13 --3-32 *<>''* 

Mole Fraction and Mole Per Cent. In a 1 mole per cent solution 
of x, 1 per cent of the number of molecules present are x molecules. 

Example 7. How would you prepare a 1 mole per cent solution of 
LiF in water? Dissolve 1 mole = 25.94 grams of LiF in 99 moles = 
99 X 18 = 1,782 grams of water. 

Example 8. How would you prepare a 2.5 mole per cent solution of 
KC1 in water? Dissolve 2.5 moles = 2.5 X 74.50 = 180.40 grams of 
KC1 in 97.5 moles = 97.5 X 18 = 1,755 grams of H 2 O. 

1 o is read "is equivalent to" and means, for instance, that 31.13 grams of 
LiF are contained in 1,031.13 grams of solution. 



96 QUALITATIVE ANALYSIS 

Example 9. What is the mole fraction and mole per cent of HC1 in 
a 2.5 molal solution? Molecular weight of HC1 = 36.46. There are 
present 2.5 moles of HC1 for every 1,000/18 grams = 55.6 moles of 
water in a 2.5 molal solution. 

2 ' 5 2 ' 5 = 0.043 = mole fraction of HC1 



2.5 + 55.6 58.1 
0.043 X 100 = 4.3 mole per cent of HC1 

Normal Solutions. A IN solution of a substance contains one 
equivalent of the substance for every 1,000 ml. of solution. 

For acids, bases, and salts, the equivalent is the molecular weight 
divided by : the valence number of the ion multiplied by the number of 
these ions present in the molecule. 

For oxidizing and reducing agents, the equivalent is the molecular 
weight divided by: the valence change of the ion acting as an oxidizing 
or reducing agent multiplied by the number of such ions present in the 
molecule. 1 

Example 10. Prepare a 1.75N solution of H 2 S0 4 . Molecular 
weigh't of H 2 SO 4 = 98. Valence number of SO 4- = 2; number of 
SO 4 ~ = 1; valence of 1I + = 1; number of H+ = 2; valence num- 
ber X no. = 2. 



= 49 g./liter of solution =c= IN solution 
1.75JV =0= 49 X 1.75 = 85.75 g./liter solution 

Example 11 Prepare a 0.0823N solution of K(SbO)C 4 H 4 6 'KH 2 O, 
potassium antimonyl tartrate (tartar emetic). Molecular weight = 
333.9. Valence number of K 4 " = 1; number of K + = 1. Valence 
number of (SbO) + = 1; number of (SbO)+ = 1. Valence number of 
(C 4 H 4 O 6 )- = 2; number of (C 4 H 4 6 )- = 1. 

X 0.0823 = 27.48 g./liter of solution 






27.48 grams per liter is needed to make an 0.0823 AT solution of tartar 
emetic with respect to either the potassium or the antimonyl ion. 

X 0.0823 = 13.74 g./liter of solution 



13.74 grams per liter is needed to make an 0.0823JV solution of tartar 
emetic with respect to the tartrate ion. 

Example 12. How many grams of salt are present in 24.7 ml. of a 
1.31N solution of K 2 IIPO 4 ? Molecular weight of K 2 HPO 4 = 174.2; 

1 See Chap. IX for a thorough discussing of oxidation and reduction. 



REACTIONS WITH SOLVENTS 97 

1 liter of a 1.312V solution contains 1.31 equivalent weights of solute. 

24 7 
24.7 ml. contains 1 nnn X 1.31 = 0.0324 equivalents of solute 

J , 



Equivalent weight of K 2 HP0 4 with respect to H+ = 174.2 

1 74 2 
with respect to K+ = ^^ = 87.1 

2t 

\ 74 2 
with respect to PO.r = ^ ' = 58.1 

o 

Hence, if the solution is 1.312V with respect to II+, 2 1.7 ml. will contain 

0.0324 X 174.2 = 5.65 g. 
If the solution is 1.312V with respect to K+, 24.7 nil. will contain 

0.0324 X87.1 = 2.82 g. 
If the solution is 1.312V with respect to PO 4 *~, 24.7 ml. will contain 

0.0324 X 58.1 = 1.88g. 

Example 13. Prepare a 1.1MT solution of KMn() 4 . Molecular 
weight = 158. As a salt, a 1.102V solution contains 

158 X 1.16 = 183.3 g./liter 

As an oxidizing agent for the reaction AlnOf ' > Mu* f+ , the change in 
valence number = 5. A 1.102V solution contains 

15 ?5 X 1.16 = 36.7 g./liter 

Normal solutions arc most often used in quantitative 1 analysis, since 
their use facilitates some of the necessary computations. 

Specific Gravity. One other method is often used to express solubil- 
ity, i.e., specific gravity. The specific gravity of a liquid at some 
temperature is the weight of a given volume of that liquid (at that 
temperature) divided by the weight of the same volume of water either 
at the same temperature or at some other reference temperature. It is 
usually written as follows: 

60F 
Sp. gr. ~op- of 30.00% HC1 = 1.1526 

This means that the weight of the IIC1 solution was taken at 60F. 
and was compared to the weight of an equal volume of water at 60F. 
The quotient is 1.1526. Or it may be specified as follows: 

9n P 
Sp. gr. ~- of 30% HC1 = 1.1492 



98 QUALITATIVE ANALYSIS 

This means that the weight of the HC1 solution was taken at 20C. and 
was compared to the weight of an equal volume of water at 4C. The 
quotient is 1.1492. Specific gravity is a very common method used in 
industry for specifying the concentrations of solutions. Tables can be 
found in the handbooks for a great number of common substances that 
convert specific gravities to percentage and various other units. 

Example 14. How many milliliters of HC1 with a specific gravity 
20 C 1 
4~CT of L1789 are needed to prepare 1,500 ml. of 0.52M HC1? From 

20 
the tables, HC1 sp. gr. -TO- = 1.1789 corresponds to 36 per cent HC1 and 

contains 424.4 grams HC1 per liter. Molecular weight HC1 = 36.46. 
Hence 



36 per cent HC1 solution is 11.6AT 

x -^ - = 67.25 ml. of the HC1 solution required 

Degree of Solubility. Thus far we have discussed the case of sub- 
stances that dissolve, but what about substances that do not dissolve? 
If the forces holding the solid together are sufficiently great, the solvent 
molecules may not be able to pull the solute particles into solution; or, 
to put it another way, by the time an ion has so disposed itself that it 
can be broken away from the solid, some ions that are already in solu- 
tion will have collided with the surface and attached themselves to it. 
In other words, the equilibrium has been definitely shifted in one 
direction, toward crystallization. As a little thought will show, this 
point of view demands that all substances be soluble to some degree, 
even if the solubility is fantastically small. 

Experimental Characteristics of Solutions. One of the criteria by 
which we may classify solutions is their conductance of an electric 
current. In this classification, the extremes are, of course, solutions 
with no resistance to the passages of an electric current and solutions 
with infinite resistance to an electric current. Solutions exhibiting 
either of these two extreme types of behavior do not exist, but all the 
intermediate types do. Roughly, we can divide solutions into two 
classes: those that* conduct very poorly and those that conduct very 
well, i.e., nonelectrolytes and electrolytes. 

Nonionic Solutions. Let us first consider, briefly, nonelectrolytes. 
Here the solution is the so-called molecular solution; i.e., the molecules 



REACTIONS WITH SOLVENTS 99 

are present as such in the solution. In dissolving a solid that forms 
solutions of this type,, the solvent usually acts against the van der 
Waals' forces that hold such a crystal together. If the solution is very 
dilute, it resembles a gas in its behavior; i.e., the solute molecules arc so 
far apart and collide with each other so infrequently (relatively speak- 
ing) that we may think of the solution as a gas having solvent molecules 
instead of just space between the solute molecules. Solutions of this 
type may be dealt with mathematically in a manner reminiscent of 
gases. If the solution becomes more concentrated, the interaction 
between solvent and solute molecules becomes more complex, and such 
solutions arc not dealt with so easily. However, we shall not consider 
the reactions of this type of solution. 

Ionic Solutions. We shall consider, rather, solutions that conduct 
the electric current; i.e., solutions in which tho solute exists as ions, 
electrically charged bodies, and solutions in which the solute exists both 
in the form of molecules and ions. If we tajce two strips of litmus 
paper, one red and one blue, and place them in a solution of this typo, 
one of three things will happen. Kither the color of the blue strip will 
become red or the color of the red strip will become blue or the color of 
both strips will remain unchanged. Furthermore, if we take the solu- 
tion that turns blue litmus red and add it to the solution that turns 
red litmus blue, after a certain amount has been added, the resulting 
solution will no longer change the color of litmus; i.e., the two solutions 
will have neutralized each other. These differences in ionic solutions 
have been known for a very long time and have been given special 
names. Solutions that turn blue litmus red are called acidic solutions; 
solutions that turn red litmus blue are called basic, or alkaline, solu- 
tions. Solutions that do neither are called neutral solutions. Acidic 
solutions will neutralize basic solutions and vice versa. 

THE MODERN THEORY OF ACIDITY AND BASICITY 

Let us see why certain substances, when dissolved in a solvent, give 
rise to acidic solutions, why certain others give rise to basic solutions, 
whereas still others yield neutral solutions. 

Aqueous Solutions. Let us take an actual example. If we dis- 
solved some sulfur trioxide, SO 3 , which is a solid, in water, the sulfur 
trioxide would dissolve and, because of the intermolecular forces, would 
react with water molecules thus: 

3H 2 + S0 3 - 2H 3 0+ + SOr 

(Of course, we are not showing the existence of the whole atmosphere of 
water molecules surrounding each ion in this case.) If we tested this 
solution, we should find that it was acidic. 



100 QUALITATIVE ANALYSIS 

Let us take another example. If we dissolved some metallic 
sodium, Na, in water, the sodium would dissolve, and the reaction 
would occur as follows: 

2Na + 2/H 2 -> 2Na(H,O),+ + 2 OH" + H 2 T 

Testing this solution with litmus would show that it is basic. 

Let us now take these two solutions, one basic and the other acidic, 
and heat them, thereby evaporating the excess water. After a while, 
in the case of the basic solution, we should have left a white solid, 
whereas in the case of the acidic solution, we should have an oily 
liquid. By analysis, we could show that these substances have a 
composition corresponding to the formulas NaOH and H 2 S()4, respec- 
tively. These substances are evidently different from the parent 
compounds. However, these compounds when dissolved in water 
give solutions that are basic and acidic, just like the original solutions. 
Let us now add one solution to the other until the resulting solution 
is neutral and then evaporate the water. By this process, we obtain 
a white solid, which, on analysis, is found to have the formula Na 2 S0 4 . 
This substance, on re-solution, gives a solution neutral to litmus. 

At this point, the student probably wonders how all this is to be 
explained. Let us set down again all the reactions that have been 
studied. 

Acidic solution: 

S0 3 + 3H 2 -> 2H 3 0+ + SOr 
2H 3 + + SCV + Z6-H 2 -^ xsH 2 | + H 2 S0 4 

H 2 S0 4 + 2H 2 -* 2H 3 0+ + S0 4 ~ 
Basic solution : 

2Na + ?/H 2 -> 2Na(H 2 0) z + + 20H- + H 2 | 
Na(H 2 O) x + + OH- + xsH 2 O -^ xsH 2 O | + NaOH 

NaOH + 2/H 2 - Na(H 2 0), + + OH~ 



For the neutralization, the reaction would be: 

[2Na(H 2 0),+ + 20H-] + [2H 3 0+ + SOr] -> 

2Na(H 2 0) a5 + + SOr + 4H 2 
Neutral solution: 

2Na(H 2 0),+ + SOr + 4H 2 -^ Na 2 S0 4 + a*H 2 T 
Na 2 SO 4 + zH 2 O -> 2Na(H 2 O) I + + SOr 

Let us now examine these equations. What differences exist 
between the neutral and the acidic and the basic solutions? What has 



REACTIONS WITH SOLVENTS 101 

happened in the course of the neutralization? The most important 
thing that has happened is that the hydronium ion H 3 O+ has combined 
with the hydroxyl ion, OH~~, and reformed some molecules of solvent. 
If we examine the neutralization of many different types of aqueous 
acidic and basic solutions, we shall find that they are all accompanied 
by the formation of water molecules from the hydroxyl and hydronium 
ions present. We may say, therefore, that in aqueous solutions, the 
presence of hydronium ions will give the solution an acidic character, 
whereas the presence of hydroxyl ions will give it a basic character. 

Liquid Ammonia Solutions. Let us now consider the same phenom- 
ena in a nonaqueous solvent, e.g., liquid ammonia. Liquid ammonia is 
colorless and boils at 33.40C. at atmospheric pressure. If we 
dissolved some sodium in this solvent, the reaction would be 
2Na + T/NH 3 -> 2Na(NH 3 ) z + + 2NH 2 - + H 2 1 

and the solution would be basic. Let us make another solution by dis- 
solving H 2 SO 4 , secondary hydrogen sulfate, in liquid ammonia. The 
reaction would be 

H 2 S0 4 + 2NH 3 -> 2NH 4 + + SOr 
and the solution would be acidic. 

These two solutions can be added to one another (in the presence 
of an indicator to indicate the point of neutrality), and they will 
neutralize one another. The reaction would be 

[2Na(NH 3 ) x + + 2NH 2 -] + [2NH 4 + + S0 4 ~] -> 

2Na(NH 3 ) x + + S0 4 - + 4NH 3 

Again, as in the case of aqueous solutions, neutralization has been 
accomplished by the reforming of neutral solvent molecules from two 
ionic fragments. 

If we took the original acidic solution and evaporated the solvent, 
we should get a substance (N1I 4 ) 2 SO 4 , ammonium sulfate, which would 
give an acidic solution upon re-solution in liquid ammonia. 



2NH 4 + + S0 4 ~ + o*NH 3 -+ (NH 4 ) 2 S0 4 



(NH 4 ) 2 SO 4 + o*NH 3 - 2NH 4 + + SO 4 - 

The basic solution would also give a solid upon evaporation that, when 
redissolved, would give a basic solution. 



Na(NH 3 ) x + + NH 2 ~ + ssNH 8 -> NaNH 2 

A sodamide 



NaNH 2 + a*NH 3 -> Na(NH 3 ) x + + NH 2 - 

Selenium Oxychloride Solutions. Let us now choose another 
solvent, selenium oxychloride, SeOCl 2 . This solvent is a liquid that is 



102 QUALITATIVE ANALYSIS 

very reactive, attacking most substances. It is inert to platinum, 
glass, and tungsten. Note, however, that no hydrogen exists in this 
solvent. Let us dissolve some sulfur trioxide in selenium oxychloride. 
The reaction would be 

S0 3 + 2SeOCl 2 -> [(SeOCl 2 )(SeOCl)]+ + S0 3 C1- 

Let us also prepare a solution by dissolving sodium in selenium 
oxychloride. The reaction would be 

6Na + 4SeOCl 2 -> 6Na+ + 6Q- + 2Se0 2 + Se 2 Cl 2 



If we added one of these solutions to the other, we should find that at 
some point neutrality exists (although here we would follow the course 
of the reaction electrically by means of conductance rather than with 
an indicator). The reaction for this neutralization would be written 
thus: 



{S0 3 C1- + [(SeOCl 2 )(SeOCl)]+} + {Na+ + 01- 

S0 3 C1- 



Definition of Acidic and Basic Solutions. Let us now set down the 
neutralization reactions in each of the three systems that we have 
examined: 

Water system: 

(2Na+ + 20H-) + (2H 3 0+ + SOr) -> 2Na+ + SOr + aH,0 
Ammonia system: 

(2Na+ + 2NHr) + (2NH 4 + + S0 4 ~) -> 2Na+ + 804=" + aNH, 
Selenium oxychloride system: 

{Na+ + 01-} + {S0 3 C1~ + [(SeOCl 2 )(SeOCl)]+} -> 

Na+ + S0 3 C1- + rwSeOCli 

We can see that these reactions resemble each other; in fact, the paral- 
lelism existing is quite remarkable. For each solvent, the acidic 
solution contains a negative ion of some sort plus a positive ion that 
came from the solvent, whereas the basic solution contains a positive 
ion of some sort plus a negative ion that came from the solvent. Dur- 
ing neutralization, the negative and positive ions that came from the 
solvent recombined to form more solvent, leaving the other ions 
unchanged in solution. The ions that came from the solvent may be 
called solvo-ions; i.e., in the case of acidic aqueous solutions, the solvo- 
ion is the hydronium ion; in the case of the basic liquid ammonia 
solutions, the solvo-ion is the amido ion NH 2 ~. Hence one can say 



REACTIONS WITH SOLVENTS 103 

that an acidic solution is one that contains as one of the predominant 
species present a positive solvo-ion, and a basic solution is one that 
contains as one of the predominant species present a negative solvo- 
ion. During neutralization, the negative solvo-ions combine with the 
positive solvo-ions to form neutral solvent molecules. Hence a neutral 
solution is one in which none of the predominant species present is a 
solvo-ion. 

Limited Definition for Water Solutions. Since most elementary 
work deals only with water solutions, we can present a version of this 
definition that applies only to aqueous solutions. In the water system, 
the positive solvo-ion is the hydronium ion, and the negative solvo-ion 
is the hydroxyl ion; hence in a water solution, if the hydronium ion 
H 3 O+ is one of the predominant species present, the solution is acidic ; 
if the hydroxyl ion OH~ is one of the predominant species present, the 
solution is basic. 

NOTE: The reader may have noticed that we avoid the use of the words acid 
and base, preferring to speak about acidity and basicity. By doing this, we think 
that a rather large controversy as to^ what to call an acid and what to call a base 
can be avoided. Confining the definitions to solutions also has its advantages. 
If the current theories of acids and basis are followed, through the theories of 
Bronsted-Lowry, G. N. Lewis and Usanowitch, one discovers that in this way the 
subject of acids arid bases is so expanded as to include practically the whole of 
chemical reaction. Hence the need for choosing a halting point is apparent. 

If we deal historically with the problem, we find that acidity and basicity are 
best known in solution and most commonly used in connection with solutions, 
hence it seems wise to call a halt there. Furthermore, if we discard the nomen- 
clature^ acid and base, we find that the problem is immensely simplified, and the 
fact that both zinc chloride and hydrogen chloride give acidic solutions is not too 
difficult to understand. Similarly, the facts regarding the acidity arid basicity of 
salt solutions become clear. The only drawback in the whole picture is the name 
that we now give to hydrogen compounds, calling hydrogen chloride in water solu- 
tion hydrochloric acid and anhydrous secondary hydrogen sulfate, sulfuric acid. 
We hope, however, that in time the anachronism will disappear, as have the 
early types of nomenclature. In passing, may we state that although in this 
chapter we have adhered strictly to the policy of not using the words acid and 
base, we have not been so strict in other parts of this book, since we believe that 
it is always wise to make haste slowly, i.e., not to be too dogmatic about intro- 
ducing changes. 

Substances That Give Acidic, Basic, and Neutral Solutions. The 

logical question now is, can we predict whether a certain substance will 
give an acidic, basic, or neutral solution? And as a corollary, how 
acidic or basic will the solution be? We shall confine our discussion 
here to aqueous solutions, since they are the ones most frequently 
encountered; however, the principles developed may readily be 
extended to include nonaqueous solutions. 



104 QUALITATIVE ANALYSIS 

Let us again review our definition of an acidic and of a basic solu- 
tion. In a water solution, if the hydronium ion H 3 0+ is one of the 
predominant species present, the solution is acidic; if the hydroxyl ion 
OH" is one of the predominant species present, the solution is basic. 
There are several principles embodied in this definition. 

1. The substance must be soluble at least to some extent. This state- 
ment is fairly obvious, since a substance must dissolve if we are to have 
a solution. 

2. The substance must react with the solvent, the reaction involving 
some primary valence bonds and yielding ions. 

3. One of the predominant species present as the result of the reaction 
must be the hydronium ion if the solution is to be acidic or the hydroxyl ion 
if the solution is to be basic. 

Let us now examine these criteria further and see what the under- 
lying reasons are for these differences in behavior. A substance may 
dissolve for one of two reasons. (1) The substance will react with the 
solvent, thereby involving the secondary valence forces. This type of 
reaction is a sort of addition reaction qjid will not give rise to ions. Or 
(2) the substance may react involving the primary valence forces and 
liberating ions. Let us examine further this second case. Notice 
that the statement was conditional, "The substance may react." 
What, exactly, determines if it will react? The determining factor is 
the bond strength, i.e., the magnitude of the force of attraction between 
the atoms. Let us choose several actual cases and see how this works. 

BOND STRENGTH 

Oxy Compounds. Let us choose first sodium oxide, Na 2 0, which is 
quite soluble. The attraction between the sodium and the oxygen 
atoms is not great, and the energy required to separate them is not very 
great. Hence water can react with this " molecule " and separate the 
ions thus: 



Na:O:Na + zsH:0:H ->'2Na(H 2 0).+ + 2[:O:H]~ 

Since this reaction furnishes OH~ ions, the solution will be basic. Let 
us compare the reaction of this compound with silicon dioxide. In 
silicon dioxide, the attraction between the silicon and oxygen is very 
great, and hence it is difficult to separate the ions. Therefore Si0 2 is 
insoluble. On the other hand, in sulfur trioxide, S0 3 (which is a white 
solid), the oxygen is very strongly bound to the sulfur, and the energy 
required to separate them is very great. In fact, the sulfur has such 
an avidity for oxygen that it will break up the water and remove oxygen 
from it as 



REACTIONS WITH SOLVENTS 105 



3H:o:n- 



.0* < 
. - 



:o: 
: b : s : b : 



2H 8 O+ 



Since this reaction furnishes hydronium ions, the solution is acidic. In 
these three examples, we have examined a few members of a horizontal 
line or period in the atomic table, and we have seen that, although small 
energies are required to separate the atoms in the first groups, as we 
proceed to the right, the elements not only will not give up their part- 
ners readily but will, in fact, break up other compounds so that they 
can share more electrons. 

One other factor (except the group of the periodic table) influences 
the bond strength. This is the period in which the element is placed, 
or, to put it another way, the atomic volume. As the atomic number 
increases, the number of shells surrounding the atoms increases, or the 
volume of the atoms increases. Since the binding forces are electrical, 
they depend upon the magnitude of the charges and the distance. This 
point is illustrated by the fact that, whereas beryllium oxide, BeO, is 
insoluble and magnesium oxide, MgO, is slightly soluble, calcium oxide 
reacts completely with water, forming calcium hydroxide. These 
rules do not, unfortunately, hold exactly in the later periods; other 
factors enter that complicate the picture. 

Hydro Compounds. The hydrides are another and simpler exam- 
ple of the influence of bond strength. When sodium hydride is dis- 
solved, the sodium ion and the hydride ion are separated, and the 
latter reacts with the water to give hydrogen molecules and hydroxyl 
ions, resulting in a basic solution. 

Step 1: Na:H -^ Na+ + (H:)~ 

Step2: (H:)- + H:0:H->H:H f + [:O:H]~ 



Over-all reaction: Na:H + H:0:H->H:H | + Na+ + [:0:H]- 

At the other end of this period, we have hydrogen chloride, HCI. This 
compound reacts with water, and the electrons are attached not to the 
hydrogen but to the chlorine; so we get a hydronium ion and a chloride 
ion, giving an acidic solution. 



H:C1: +H:0:H-*[:Cl:]- 
In between these two, we have methane, CEU, in which the bond 



106 QUALITATIVE ANALYSIS 

strength is high, and accordingly, this compound is not appreciably 
soluble in water. 

Hydroxy Compounds. Here we have an interesting case. Let us 
take a hypothetical compound, ROII, which could react in one of three 
ways if added to water. 

1. The bond between the oxygen and the element R could split, 
giving a basic solution. 

ROH + H 2 -> RCHoO)^ + OH- 

This would happen if the strength of the bond between the R and the 
was weaker than the bond between the () and the IT. An actual exarn- 
- pie of this is sodium hydroxide, NaOTT. 

NaOH + zH 2 -> Na(H 2 0),+ + OH~ 

2. The bond between the oxygen and hydrogen could break, giving 
an acidic solution. 

ROH + H 2 -> RO- + H 3 i- 

This would happen if the Oil bond were weaker than the RO bond. 
An actual example of this is hypochlorous acid, HOC1. 
HOC1 + H 2 - II 3 O+ + CIO- 

3. The third alternative is that it will not react at all. 

ROH + H 2 O ^ ROH + H 2 O 

An example of this is methyl alcohol, CH 3 OIL 

Amphoteric Substances. In addition to these three types of 
behavior, there are many other substances that can react in both 
ways; i.e., either the Oil or the OR bond can be broken. 



ROII + H 2 o ( 

Ml(H 2 0).+ + OH- 

In these cases, exactly which type of behavior will occur will depend 
entirely upon what other ions are present in the solution. Stannous 
hydroxide, Sn(OH) 2 , which is not very soluble in water, is a substance 
of this sort. If we add stannous hydroxide to a solution containing 
hydronium ions, the SnO bond will break, but if we add it to a solution 
containing hydroxyl ions, the OH bond will break. 

Sn(OH) 2 + 2H 3 0+ ^^ g n ++ + 4H2 o 

JTSrliO 

Sn(OH) 2 + 20H- " H ~> SnO 2 ~ + 2H 2 O 

JTsHaO 

This particular reaction is utilized in the laboratory in order to prepare 
sodium stannite solution. Sodium hydroxide is added dropwise to a 



REACTIONS WITH SOLVENTS 



107 



stannous chloride solution until stannous hydroxide precipitates. 
Then an excess of sodium hydroxide is added, and the stannous hydrox- 
ide dissolves, forming sodium stannite. Since many of the ions that are 
encountered in the course of an analysis are distinctly amphoteric, 
these properties are often utilized to effect separations. As an exam- 
ple, the separation of aluminum from iron may be cited. If we treat a 
mixture of aluminum and ferric ions with sodium hydroxide, the first 
reaction is the separation of the hydroxides. 

Fe ++ + + A1+++ + 60H- + H 2 () -> Fe(OII) 3 (H 2 0), + Al(OH) 3 (H 2 0) tf 

If now we treat these hydroxides with an excess of sodium hydroxide, 
the aluminum dissolves, forming the aluminate ion and leaving ferric 
hydroxide behind. 

Fe(OH) 3 (H 2 O) x + xsOll~ -> no reaction 
A1(OH) 3 (H 2 ()) V + jsOII- -+ MO,- + .TKOH- + xH,0 

It is interesting to investigate the probable mechanism of this latter 
reaction. Let us start with the triply positive aluminum ion and add 
hydroxyl ions and see what happens. 



Al(H 2 O) a -H- + oil- 
Or, using the electronic symbolism 



A1(OH)(H 2 ()) 5 ++ 



H 2 



*H H' 
H:O:H 



I H I 



M 



H" 
H 

ft'* 

H 



Al 



H:O:H 



" 

*'H 
.H 

?' 
H" 



-4- H:O:H 



The next steps would be 

Al(OH)(H a O) 5 ++ + OH 

A1(OH) 2 (H 2 0) 4 + + OH- 

Al(OH)a(HaO)j + OH- -> A1(OH) 4 (H 2 ()) 2 - 

A1(OH) 4 (H 2 O) 2 ~ is equivalent to A1O 2 (H 2 O) 4 



Al(OH)a(H 2 <))a 



A1(OII) 2 (H 2 0) 4 + + II 2 O 

+ H 2 O precipitate 
+ H 2 O 

, which is the meta alumi- 
nate ion 



Beautiful as this mechanism looks, however, the student should not be 
misled into thinking that it represents the whole truth. It really does 
not. It represents certain aspects of it. Let us examine the facts arid 
see what we can gather from them. From analytical data, we know 
that aluminum chloride usually crystallizes as the hexahydrate and 
accordingly has the formula A1C1 3 -6II 2 O. This has been interpreted 
to mean that the aluminum ion coordinates six atoms of water and that, 



108 



QUALITATIVE ANALYSIS 



the chloride ions are held by heteropolar bonds. Hence the symbolic 
picture would be 



f - 

9*' '# 

H .. H 

H:O:H 



4- 3CP 



When the hcxaquo-aluminum ion reacts with a hydroxyl ion, one of the 
hydrogens is removed, and the picture is 



r H 



H ^ 


H. 


o. 


*o** 


' A1 


H 


O* 


:o*. 


H H:O:H 


H 



However, the oxygen, which has only one hydrogen atom attached to 
it, is not located uniquely, and an interchange or migration of hydrogen 
atoms may occur. If we remove two hydrogen atoms by reaction 
with two hydroxyl ions, we can draw two alternate pictures. 



H H:O:H 


t H 




? H:(J:H H 


f Al 


_ H 


or 


*' Al '^ 
H A1 H 


*H 


'& 




:6; $. 


H 


_ 




: ?. : 



Any two of the oxygen atoms may have only one H each. 

Either of those can exist, and both probably do exist in equilibrium 
with each other. Similarly, if we remove four hydrogen atoms, we 
have the following pictures: 



H H: ?. :H H 



Al 



f\9 

H 'S* H 



Al 



or Al or H Al 

^ H:O:H ^ 



H:O:H 



All of these probably exist in equilibrium with each other. In solution, 



REACTIONS WITH SOLVENTS 109 

the structure [Al(OH)4(HjO) J~ probably predominates. If, however, 
one dehydrates the solution by evaporation of the water, since the 
form [A10 2 (H 2 0)4]"" can lose the most water, the equilibrium is shifted 
in that direction, and the final dry product is NaA10 2 ; hence the com- 
mon formulation of the aluminate ion as A10f . 

General Conclusions. We thus see that we can predict in the case 
of hydro, hydroxy, and oxy compounds of the elements what reaction 
with H 2 O will take place. The general prediction will be that com- 
pounds whose central atoms are in the first groups of the periodic table 
will be basic, whereas those in the last groups will be acidic, and those 
in the intermediate groups will be cither inert or amphoteric. 

SOLUTIONS OF SALTS 

It is also of interest to see whether we can predict whether a solu- 
tion of a salt will be acidic, basic, or neutral. The answer is that to 
some extent we can. Let us take some examples: 

1. We wish to predict what sort of solution sodium chloride will 
give when dissolved in water. The first reaction would be the separa- 
tion of the ions. 

Na+Cl- + o:sH 2 O -> Na(HiO),+ + CI- 

As we know from previous discussions and reactions, the sodium ion 
does not react with water (except to form a complex ion). Similarly, 
the chloride ion does not react with water. Hence the solution would 
be neutral. 

2. We wish to predict what sort of solution sodium cyanide, 
Na + CN~, will give. First let us separate the ions. 

Na+CN- + xsII 2 O -> Na(H 2 0) x + + CN~ 

The sodium ion does not react with water, but the cyanide ion does; 
thus 

ON- + H 2 -> HCN + OH- 

Hcnce the solution will be basic. 

3. As a third illustration, we wish to predict what sort of solution 
we shall get with ammonium chloride, NH 4 C1. 



NH 4 +C1- + zsH 2 -> NH 4 + + Cl- 
The chloride ion does not react with water, but the ammonium ion does. 

NH 4 + + H 2 -> NH 3 + H 8 0+ 
Hence the solution would be acidic. 



110 QUALITATIVE ANALYSIS 

4. Let us take still another example, ammonium acetate, 
NH 4 C2H 3 O 2 . 

NH 4 +C 2 H 3 2 - + xsH*0 -> NH 4 + + C 2 H 3 O 2 - + aH,0 
Here both ions react with water. 

(a) NH 4 + + H 2 -> NH 3 + H 3 0+ 
(6) C 2 H 3 2 - + H 2 O -* HC 2 H 3 O 2 + OH~ 
and 

(some OH~~ or 

H 3 0+ + OH~ - H 2 + {some H,0+ or 

( neither 

Whether the final solution will be weakly acidic or weakly basic or 
neutral will depend entirely upon which reaction, (a) or (6), goes further 
to completion. Tn other words, it will depend upon the equilibrium 
constants. Let us write these equilibrium constants. 

For (a) 

_ [NHd[H,0+] 

* " [NH 4 + TlH 2 0] 

where [NH 3 ] = concentration of ammonia, etc. 

or, since the concentration of the water is essentially constant 



[NH,][H,0+] r , 
K = "" = [ 2 ] 



For (6) 



~ [C 2 H 3 () 2 -][H 2 or 
or 

jff _ [HC 2 H 3 O 2 ][OH-] 
/v --- ?->,- w v^-zi ~~ K L^uj 
[U 2 H 3 O 2 J 

It can be seen from these equations that if the reaction in equation (a) 
goes further to the right than does reaction (6), K will be greater than 
K'j and hence the solution will be acidic and vice versa. In this par- 
ticular case, K and K' are approximately equal, and the solution is 
neutral. 

Therefore, in the case of the simpler salts, we may list four cases: 

I. Neither the cation nor the anion reacts with water. The solu- 
tion therefore is neutral. 

II. The anion but not the cation reacts with water. The solution 
is basic. 

NOTE: If the substance being dissolved already possesses hydroxyl ions in its 
structure and these ions are separated by the water, the solution will be basic, 
since there is thus an increase in the number of hydroxyl ions present. 



REACTIONS WITH SOLVENTS 111 

Example: 



NaOH + *sH 2 -> Na(H,0),+ + OH~ 

III. The cation but not the anion reacts with water. The solution 
is acidic. 

NOTE: If the substance being dissolved already possesses hydronium ions in 
its structure and these ions are separated by the water, the solution will bo 
acidic, since there is thus an increase in the number of hydronium ions present. 
Example: 

H 2 S0 4 -2H 2 EEE (H 3 0) 2 S0 4 

(H 3 0) 2 S0 4 + xsll.0 -> 2H 3 (H + SOr + roHjO 

IV. If both the cation and the anion react with water, the solution 
may be acidic, basic, or neutral, depending upon which reaction goes 
further to completion. 

Reactions of this type, i.e., reactions of salts witli water, arc 
generally called hydrolytic reactions, or hydrolysis. 

More Complicated Cases of Salt Solutions. We have thus far 
considered anions and cations that react with water in one way, but 
there are ions that can react .in more than one way, e.g., the ortho- 
phosphate ions. Let us consider first hydrogen phosphate, I1 3 PO 4 , and 
its water solution. The first reaction is 

HP0 4 + H 2 ^ H 3 0+ + H 2 P0 4 - (1) 

Secondary 
oithophos- 
phate ion 

and the solution is acid. The equilibrium constant for this reaction is 

[H,0'][H 2 ]>q 4 -] 
"' 



WHO! K [H.O+][H,POr] _ . , 
*[H,0] = K = > ~ ' 



The H2P04~~ ion can react further in two ways: 

H 3 0+ + HPOr (2) 

# 

H 2 P0 4 - + H 2 

^ 

H 3 P0 4 + OH- (3) 

However, reaction (3) does not go appreciably from the left to the right 
but, rather, from the right to the left. Reaction (2) proceeds to some 
extent. The equilibrium constants are 

K , _ [H 3 0+][HP0 4 =] on v 10-7 

K - "P07T~ 2 ' x 

= - 91 * 10 " 12 



112 QUALITATIVE ANALYSIS 

K' is greater than K"\ therefore reaction (2) predominates. 
The HP0 4 "" ion can react further; thus 

H 8 0+ + P0 4 S (4) 

^ 
HP0 4 ~ + H 2 

* 

H 2 P0 4 - + OH- (5) 

Reaction (5) is predominant, as may be seen from the equilibrium 
constants 

r ,,, _ [H 3 0+][POr] 
* ~ [HPOr] ~ x 
,, _ [OH-][H,P0 4 -1 _ r n v m-8 
K -- [HPO?] -- 5 ' X 10 

and K"" is greater than K'". 

The final ion PQF can react in only one way. 

PO^ + H 2 ^ HPOr + OH- (6) 

and the equilibrium constant is 

vi nil [OH~][HP04~] o 7C s^ 1 n-2 

K -- [PO?J 2 - 78 x 10 

Let us see how this may be applied to the phosphate salts. If we 
dissolve primary sodium orthophosphate, NaH 2 PC>4, in water, what 
sort of solution will wo get? 

NaH 2 PO 4 + o*H 2 O -> Na(H 2 O) x + + H 2 PO 4 - 

H 2 POr + H 2 5= H 3 0+ + HP0 4 3 (7) 

This reaction is the same as (2) and would predominate; hence the 
solution would be acid. If we dissolve secondary sodium orthophos- 
phate, Na 2 HP0 4 , in water, the reaction will be 

Na 2 HPO 4 + zsH 2 O -> 2Na(H 2 O) x + + HPOr 

H 2 POr + OH- (8) 

HP0 4 - + H 2 

^ P0 4 - + H 3 0+ (9) 

Since (8) is the predominant reaction, as can be seen from the equilib- 
rium constant, the solution will be basic. If we dissolve tertiary 
sodium orthophosphate, Na 3 PO 4 , in water, the reaction will be 



Na 8 P0 4 f! 3Na(HsO),+ + P0 4 a 

PO 4 - + H 2 O ^ HPOr + OH- (10) 



REACTIONS WITH SOLVENTS 



113 



and 



HP0 4 - + H 2 O 



r + OH~ 



(11) 



Hence the solution would be basic and more basic than the solution of 
secondary sodium orthophosphate, since there are more hydroxyl ions 
formed. 

TABLE XV. IONIZATION CONSTANTS 





K 


Degrees 
centigrade 


HC 2 H 3 O 2 


+ H 2 ^ H 3 0+ + 


c,H,or 


1.8 X 10-* 


20 


H 2 C0 3 


+ H 2 O ^ H 3 O+ -f 


HCO,- 


4.3 X 10~ 7 


20 


HCOr 


4- H 2 O ^ H 3 0+ + 


cor 


4.7 X 10-' 1 


20 


HCN 


-f H 2 O ^ H 3 + + 


CN" 


7 2 X 10- 10 


25 


HNO 2 


+ H 2 O ^ II 3 O+ + 


NO 2 - 


4 5 X 10~ 4 


20 


H 3 P0 4 


+ H 2 ^ H 3 + + 


H,por 


I.I X 10~ 2 


18 


H 2 POr 


+ H 2 ^ H 3 0+ + 


HPOr 


2 X 10- 7 


18 


IIPO 4 - 


+ H 2 ^ H 3 0+ + 


p() 4 ^ 


3.6 X 10-" 


18 


H 2 S 


+ H 2 O ^= H 3 O+ + 


HS~ 


9 X lO- 8 


25 


HS~ 


+ H 2 ^ H,0 + + 


s- 


1.2 X 10- 16 


25 


H 2 S0 3 


+ H 2 O ^ H.O+ + 


HS0 3 - 


1.7 X lO- 2 


25 


HSOr 


+ H 2 ^ ![/)+ + 


so 3 - 


5 X 10- 


25 


HC10 


+ H 2 ^ H 3 0+ + 


cio- 


3.7 X 10- 


17 


HN 3 


+ H 2 ^ H 3 + + 


Nr 


1.9 X 10-6 


25 


NH 3 


+ H 2 O ^ NH 4 + + 


OH- 


1.75 X 10- 


20 


Zn(OII), - 


H6II 2 O^Zn(H 2 O) 


6 ' * + 2OII- 


1.5 X 10~ 9 


25 



Another type of reaction is that given by zinc chloride, which, upon 
solution, yields 

ZnCl 2 + aH 2 O - Zn(H 2 0) 6 ++ + 2C1~ 

The chloride ion does not react appreciably with water, but the hexa- 
aquozinc ion does. 

Zn(H 2 0) 6 ++ + H 2 O ^ [Zn(OH)(H 2 0) 6 ] + + H,O+ 

Hence the solution is acid. Many other salts act in the same manner, 
e.g., aluminum chloride, A1C1 3 . 

AlCli + awHiO ~> Al(HiO)-H-*- + 3C1- 
A1(H 2 0) 6 +++ + H 2 ^ [A1(H,0) B (OH)]-H- + H 3 0+ 

If we gradually remove the hydronium ions by adding hydroxyl ions, 
the reaction is driven to the right in an attempt to maintain the equilib- 
rium, and after a while, insoluble A1(OH) 3 (H 2 O) 3 is precipitated. 



. [A1(H 2 0) 5 (OH)]++ 
. [A1(H 2 0) 4 (OH) 2 ]+ 



H 2 - [A1(H 2 Q) 4 (OH) 2 ]+ + H 3 0+ 
H 2 ^ [A1(OH) 3 (H 2 0) 3 ] + H 8 + 



114 QUALITATIVE ANALYSIS 

In many cases, it is not even necessary to remove the hydronium ions 
by adding hydroxyl ions but is necessary only to reduce its concentra- 
tion by dilution, as in the case of bismuth trichloride, BiCl 3 . 

BiCl 3 + H 2 - Bi(HjO).+++ + 3C1- 
Bi(H 2 0) x +++ + H 2 () -> [Bi(OH)(H,0)_i]++ + H 3 0+ 
[Bi(OH)(H 2 0) I _ 1 ]++ + H 2 - [Bi(OH) 2 (H 2 0) x _ 2 ]+ + H 3 0+ 
[Bi(OH) 2 (H 2 0)s_ 2 ]+ + Cl--[Bi(OH) 2 Cl(H 2 0)J 
[BifOHJaCKHiO),,] - BiOCl J, + zH 2 

Bismuth 
oxychloride 

Degree of Acidity or Basicity. The second question that we must 
answer is how acidic or basic is the solution? In other words, what are 
the concentrations of hydronium and hydroxyl ions? We know that 
hydroniurn and hydroxyl ions will react to form water 

H 3 0+ + OH- -> 2H 2 

but we can also ask, is the reverse reaction possible? The answer, of 
course,* is that all reactions are reversible to some extent. The ques- 
tion then becomes, how reversible is the reaction, or what concentra- 
tions of hydronium ions and hydroxyl ions exist in equilibrium with 
pure water? To answer this, let us write the equilibrium constant for 
the reaction 

2H 2 ^ H 3 0+ + OH- 
* - [H*0 + ][OH-] 

[H 2 0] 2 

and since the reaction goes forward only to a minute extent, the con- 
centration of the water is practically unity, and the equilibrium, or 
ionization constant, as it is often called, becomes 

K w = [H 3 + ][OH~] 

The value of this constant has been determined experimentally at 
various temperatures. These values are shown in Table XVI. Since 
many calculations arc carried out for room temperature, the value for 
25 is commonly used, i.e., 1.008 X 10~ 14 , or approximately 1 X 10~ 14 . 
Since the concentration of the hydronium ion is equal to the hydroxyl 
ion and since their product is equal to 1 X 10~ 14 , it is evident that the 
concentration of the hydronium ion (as well as the concentration of 
the hydroxyl ion) in pure water at 25 is the square root of 10~ 14 
or lO" 7 . 

[H 3 0+][OH~] = 10~ 14 

[H 8 0+] = [OH-] = 10- 7 



REACTIONS WITH SOLVENTS 115 

TABLE XVI. VARIATION OF IONIZATION CONSTANT OF WATER WITH TEMPERATURE 

(H. S. Harned and W. J. Hamer, 1933) 
Temperature, 
Degrees Centigrade K w 

0.11 X lO" 14 

10 0.29 X 10-' 4 

25 1.01 X Mr 14 

40 2.92X10- 14 

50 5.48 X 10- 14 

100 (56) X 10~ 14 

This symbol is often a little awkward to handle, and a simpler method 
of representing these facts has been devised. Let us take the common 
logarithm of both sides of this equation. 

[H 3 0+] = 10- 7 
log[H 3 + ] = -7 log 10 
and since 

log 10 = 1 
log [H 3 0+] = -7 

multiplying both sides by 1 gives 

- log [H.O+] - 7 

Let us now replace the quantity log [H 3 () f ] by the symbol pII 3 O or, 
more simply, pH. The equation becomes 

pH = 7 

That is, the pll of pure water at 25C. is 7. If the pi I of a solution is 
less than 7, the solution is acidic; if the pi I is greater than 7, the solution 
is basic. Let us illustrate with a few examples. 

Example 1. The pit of a solution is 3. Is the solution acidic or 

basic? 

pH = 3 
- log [H 3 0+] = 3 

log [H 3 0+] = -3 log 10 

[H 3 0+] = 10- 3 
if 

K w = [H 3 0+][OH-] = 10-" 

(10- 3 )[OH~] = 10- 14 
irv-14 
[OH-] = ^ = 10- 

Since 10~ 3 is greater than 10~ H , the hydronium ion is the predominant 
species, and the solution is acidic. In the preceding case 

pOH = 11 
i.e. 

- log [OH~] = 11 = pOH 



116 QUALITATIVE ANALYSIS 

Example 2. The pH of a solution is 9. Is the solution acidic or 
basic? 

pH - 9 = - log [H 8 0+] 
[H 3 O+] = 10- 9 

K w = [OH-](10- 9 ) = 10- 14 

[OH-] = 

10~ 9 is less than 10~ 6 ; therefore the hydroxyl ion is predominant, and 
the solution is basic. The problem can be simplified by applying the 
laws of logarithms. 

K w = [H 8 0+][OH-] 

and taking the logarithms of both sides 

log K w = log [H 3 0+] + log [OH-] 
or 

- log K* = - log [H 3 0+] - log [OH-] 
since 

- log [H 3 0+] = pH 
- log [OH-] = pOH 
and 

- log K w = pK w 

pK w = pH + pOH 
since 

K w = 10- 14 

- log K w = 14 log 10 = 14 = pK w 
or 

14 = pH + pOH 

Hence if we have a solution of pH = 2, the pOH = 12; or if the solu- 
tion has a pH = 1, the pOH = 15, etc. If the pH is less than pOH 
(in numerical value), the solution is acidic; if it is greater than the pOH, 
the solution is basic; and if the pH = pOH, the solution is neutral. 

Let us now see how we can calculate the pH of a given solution. 
Let us take the cases that we have previously discussed on pages 109^". 

I. Neither the anion nor the cation reacts with water. The 
solution is neutral. 

NaCl + xsH 2 O -> Na(H 2 O);+ + Cl~ 

Since no additional hydronium or hydroxyl ions are added to the solu- 
tion by this reaction, the concentration of these ions in water remains 
undisturbed; i.e. 

pH = pOH = 7 

and the solution is neutral. 



REACTIONS WITH SOLVENTS 117 

II. The anion but not the cation reacts with water. The solution 
will be basic. 

NaCN + zsH 2 O -> Na(H 2 O) x + + CN~ 
CN- + H 2 ^ HCN + OH- 

_ [HCN][OH-] 
A " [CN-] 

From tables, it can be seen that the constant for the reaction 

o*H 2 O + HCN ^ H 3 O+ + CN- 
is 






= 

<-* * 1U 



[HCN] 

but since 

K a = [H 8 0+][OH-] 



Substituting, we obtain 

K" = 



[OH-][HCN] 



K = = 
' 



K' [CN-] 



7.2 X 10- 10 
K = 1.4 X 10- 8 
Now 

_ [HCNHOH-] 



The concentration of the HCN is unknown but is* equal to the concen- 
tration of OH~ and can therefore be represented by x. If the concen- 
tration of the cyanide ion originally placed in the solution was 0.01M, 
the final concentration of CN~ is (0.01 x). Therefore 



1 4 X 10-* = 
IA X 1U 



(0.01 - x) 
and 

a? = (1.4 X 10- 7 ) - (1.4 X 10- 6 )a; 
s f + (1.4 X 10- 5 )x - (1.4 X 10- r ) = 

This may be solved by the quadratic formula, giving 

x = 3.67 X 10- 4 = [OH-] 
- log [OH-] = - log 3.67 + 4 log 10 
pOH = -0.565 + 4 = 3.44 
pH = 14 - 3.44 = 10.56 

III. The cation but not the anion reacts with water. The solution 
is acidic. 



118 QUALITATIVE ANALYSIS 

0.01M solution of NH 4 C1 
NH 4 C1 + a-sH 2 O -> NH 4 + + Cl- + 
NH 4 + + zsH 2 ^ NH 3 + H 3 0+ + a;sH 2 O 
_ [NH 3 ][H 3 0+] 
[NH 4 +] 

for the reaction 

NH 3 + H 2 ^ NH 4 + + OH- 

_ [NH.+][OH-1 _ 
A [NH 3 ] ~ L75 X 1U 

[OH-] 



[H 3 + ] 
Substituting 

R" = t NH4+ J . 



[NHJ '[H 3 0+] 

^ [NH 3 ][H 3 0+] IP" 14 

X' [NH 4 +] (1.75 X 10- 6 ) 

K = 0.57 X 10- 9 = 5.7 X 1Q- 10 

jf _ 5 7 w 
A o./ x 



(5.7 X 10- 12 ) - (5.7 X 10- w )(z) = x 2 
x 2 + (5.7 X 10- 10 )(a;) - (5.7 X K>- 12 ) = 

x = (2.4 X 10-) = [H 3 0+] 
- log [H 3 0-] = - log 2.4 + 6 log 10 
pH = -0.38 + 6 = 5.62 

IV. If both the cation and the anion react with water, the solution 
may be acidic, basic, or neutral. 

Example 1. 0.01 M solution of ammonium acetate. 



NH 4 C 2 H 3 2 + zsH 2 -> NH 4 + + C 2 H 3 2 - 
NH 4 + + zsH 2 O ^ NH 3 + H 3 0+ 

_[NH 3 ][H 3 0+]_ 57 

JfVction - rvTTT ii ' A 1U 

IJN1 4 T J 

C 2 H 3 2 - + a;sH 2 ^ HC 2 H 3 2 + OH~ 
_ [HC 2 H 3 2 ][OH-] _ 

^anion r/~^\ TT /-v _i tl.U /\ 1U 

I^2il3^2 J 



x 



Since x is so small 

0.01 x = approximately 0.01 
a? 1 = 5.7 X 10~ 12 



REACTIONS WITH SOLVENTS 119 

x = [H 3 0+] = 2.38 X 10~ 6 

K . = ^' ' = 5 fi V 10- 10 
amon (0.01 - y) X U 

Similarly, since y is small 

(0.01 - y) = 0.01 

?/ 2 = 5.6 X 10- 12 
v = [OH-] = 2.4 X 10- 6 

and since the concentration of OH~ is equal (approximately) to the 
concentration of the II 3 + , the solution is neutral. 
Example 2. 0.01M* ammonium hypochlorite. 

NH 4 C10 + xsH.O -> NH 4 + + CIO" 
NH 4 + + H 2 ^ NH 3 + H 3 0+ 



_ aa _ 

^-cation nvTTT 4-1 " ' ^ 1U 

11M114^J 

x - (of^> - 5 - 7 x 10 ~ 10 

x 2 = 5.7 X 10- 12 
x = [H 3 0+] = 2.38 X 10- 6 
CIO- + H 2 ^ HC1O + OH- 
_ [OH-][HC10] _ 

* v amon [{~^\f} _ 1 

7 n v i n-8 
/<U X U 



. 
ftnion "" (0.01 - y) ~ 

?/ 2 = 27.0 X 10~ 10 
[OH-] = y = 5.19 X 10- 5 
Recapitulating 

[OH-] = 5.19 X 10~ 5 = 51.9 X 10~ 6 
[H 3 0+] = 2.38 X 10- 6 

Since they will neutralize one another, the final [OH~] will be 
[OH-] - [H 3 0+] = (51.9 X 10~ 6 ) - (2.38 X 10~ 6 ) 
[OH-] = 49.52 X 10~ 6 = 4.95 X 1Q- 5 
pOH = - log [OH-] = - log 4.95 + 5 log 10 
pOH = -0.69 + 5 = 4.31 
pH = 9.69 

The solution is basic. 

Conclusions. Unfortunately, these methods of calculation have 
their limitations. (1) They apply only to very dilute solutions, i.e., 
solutions in which the molecules are widely separated by solvent 
molecules. (2) They do not apply to all substances that give acidic or 
basic solutions; they apply only to solutions in which the hydronium 



120 QUALITATIVE ANALYSIS 

ion concentration fe relatively low. If the hydronium ion concentra- 
tion is high, other effects enter that complicate the picture immensely. 
(3) The calculations made here use the actual concentrations, whereas a 
quantity called the activity should be used. The activity is the effec- 
tive concentration of the ion in question, i.e., the concentration that it 
appears to possess. The activity is an empirical factor that is deter- 
mined experimentally. In very dilute solutions, the activity and the 
concentration approach equality; hence the usefulness of these calcula- 
tions when applied to such solutions. 

READING REFERENCES 

LUDER, W. FAY: The Electronic Theory of Acids and Bases, Chem. Rev., Vol. 27, 

December, 1940. 
SMITH, G. B. L.: Selenium Oxychloride as a Solvent, Chem. Rev., Vol. 23, August, 

1938. 
Symposium on Acids and Bases, J. Chem. Education* Mack Printing Company, 

Easton, Pa. 

QUESTIONS 

il How many grams of lead acetate are needed to prepare a liter of 1.62M 
solution, A 1.62N solution? 

2. What is the molality of a 3.10 mole per cent solution of cadmium chloride? 
What is the weight per cent composition? 

20 

3. How many milliliters of H2SO 4 sp. gr. -^ = 1.855 are needed to prepare 

125 ml. of a 0.75 AT solution? What is the molarity of this solution? 

4. What is the normality of 99% sulfuric acid? 36% HC1? 70% HN0 3 ? 

5. What is the normality of a 12% solution of K2Cr04 relative to potassium? 
To chromate ion? 

6. What mole fractions of H 2 and BeS0 4 are present in a 7.20 molal solution 
of BeS0 4 -4H 2 O? 

7. Given a 22.4% solution of microcosmic salt, what is the mole fraction of 
each ion and compound present? 

8* A 0.5M solution of K 2 Cr 2 07 is of what normality? 

9. The solvo-ions in solutions of liquid sulfur dioxide are SO*" and SOa*" 1 ". 
Thionyl chloride, SOC1 2 , gives a basic solution in this solvent, and potassium sulfite 
gives an acidic solution. Write reactions for the solution of these substances in 
SO 2 , and write the neutralization reaction. 

10. What is the pH of a solution that has a [H 8 O+] - 2.35 X 10~ 8 ? Is the 
solution acidic or basic? 

. 11. What are the [H 3 + ] and [OR-] in solutions having the following pH: 7.32, 
3.42, 1.01, -3.25, 11.2, 18, 0, 14? 

12. What is the pH of an 0.012M solution of ammonium sulfate? 

13. What is the pOH of an 0.01 1M solution of sodium azide? Is the solution 
acidic or basic? 

14. What is the pH of an 0.05M solution of NH 4 C1? Of an 0.005M solution? 
Of an 0.0005M solution? 

15. What sort of solution do you predict that the following will give in liquid 
NH,: KC1, KNH 2 , (NH 4 ) 2 S0 4 , NH 4 N0 8 , Li, S0 8 ? Write the reactions. 



CHAPTER VIII 

REACTIONS OF CHEMICAL COMPOUNDS 
III. REACTIONS IN SOLUTION 

The ability of the chemist to predict the outcome of future experi- 
ments is very likely to impress the outsider as a faculty bordering on the 
supernatural. In reality, as Sherlock Holmes would have assured his 
friend Dr. Watson, "It is all very elementary" (once you know how it 
is done). The most dramatic example is, of course, that of Mendelejeff 
predicting the discovery of a then unknown element and describing it in 
minute detail as if he had had a sample of it at hand. We shal study 
somewhat less spectacular examples of chemical forecasting. These 
more prosaic cases are nevertheless the backbone of the chemists' 
work. 

Mass-action Law. One of the important things that the chemist 
would like to be able to predict is the effect of the addition of one solu- 
tion to another. Sometimes this is possible; many times it is impossi- 
ble. Many exact predictions are based upon certain principles, one of 
which is the mass-action law. In Chap. VI, we considered the forma- 
tion and decomposition of hydrogen iodide. In this case, we saw that 
an equilibrium constant could be written for the reaction 

2HI ^ H 2 + I, 

jr __ CH-, X Cl t 
J\. fi ' 2 

CHI 

This particular equilibrium constant remains essentially constant for 
any definite temperature. Hence we could calculate the concentration 
of HI that would result from the interaction of varying proportions of 
H and 1 2. This principle can be extended to the case of solutions. 
If we have two substances, A and B, in solution, which react to form C 
and D 

A + B^C + D 

we can express the equilibrium constant thus: 

IT Cc X Cp 
C A X C B 

If into this system we introduce more of D, the quantities of the various 



122 QUALITATIVE ANALYSIS 

components present will change so that the equilibrium constant 
remains fixed; i.e., the addition of C or D will result in the formation of 
more A and B and vice versa. This principle is termed the mass- 
action law. In a great many cases, the extent of the reaction of C and 
D to form A and B is much smaller than the opposing reaction. Hence 
the forward reaction may go practically to completion. There are 
several reasons why a reaction goes practically to completion. These 
will be considered separately. 

CASE I. DIFFICULTLY SOLUBLE PRECIPITATE 
AS A REACTION PRODUCT 

We shall choose as our first example the familiar test for barium that 
consists of the addition of a soluble sulfate to a solution containing 
barium ion to form a white precipitate of barium sulfate. We must 
now take into account the equilibrium between the solid substance 
BaS0 4 and its solution. This equilibrium can be referred to as a 
" heterogenous equilibrium/' or "complex equilibrium," as some 
authors prefer to call it. Barium sulfate is soluble in water at 25C. 
to the extent of about 2.5 mg. per liter of solution or 1 part in 400,000 
parts of water by weight. Barium sulfate is, of course, a strong elec- 
trolyte. A saturated solution is so dilute that we may assume that the 
activity coefficient of the ions is unity. In applying the mass-action 
law to the equilibrium, we must remember that we are concerned with 
an equilibrium of the solid barium sulfate, with a solution of its ions : 



X Cso 4 - [Ba++][80r] 

- " 



Since the activity of a solid is constant, then 

K 8P = [Ba++][S04-] = XJBaSOJ 

The new constant is termed the solubility product constant. 

Let us consider another example, a saturated solution of lead 
chloride. Each molecule of lead chloride yields, on solution, one lead 
ion and two chloride ions, but we may write the reaction in a stepwise 
fashion thus: 



Wd ;= PbCl+ + Cl- (1) 

and 

xsUzO 

PbCl+ ;=i Pb++ + Cl- (2) 

The equilibrium expression for reaction (1) is 

_ [PbCl+][Cl-] 
1 [PbCl 2 ] 



REACTIONS IN SOLUTION 123 

and for reaction (2) is 



z ~ TPbCFf 
Multiplying the first by the secogd, we get 

K- - K v K- = [PbCl+][Cl-][Cl-][Pb++] 

f ~ * A 2 [PbCl 2 ][PbCl+] 

or 



c " [PbClJ 

Since the same reasoning applies in this case as in the case of a saturated 
solution of barium sulfate, it follows that 

K 9P = 



Definition of the Solubility-product Principle. We may now sum- 
marize the solubility-product principle. In the case of a saturated 

TABLE XVII. SOLUBILITY-PRODUCT CONSTANTS 

PbCl 2 . 1 X 10~ 4 (25C.) 

AgCl 1.1 X 10- 10 (20C.) 

H g2 Cl 2 2 X 10- 1 * (25C.) 

HgS 3 X 10- 63 

PbS 1 X 10~ 29 

CuS 4.0 X 10- 3 * 

CdS 3.6 X 10~ 29 (18C.) 

SnS 8.0 X 10~ 29 

NiS 1.4 X 10~ 24 (18C.) 

CoS 3.0 X 10- 26 (18C.) 

FeS 3.7 X 10~ 19 (18G.) 

MnS 1.4 X 10~ 15 (18C.) 

ZnS 1.2 X 10~ 23 (18C.) 

A1(OH) 3 1.9 X 10~ 33 

Cr(OH) 3 6.7 X 10~ 31 

BaCr0 4 2.0 X 1Q- 1 " (25C.) 

SrS0 4 2.8 X 10~ 7 (17C.) 

CaC 2 O 4 2.3 X 10~ 9 

MgNH 4 PO 4 2.5 X 10~ 13 (25C.) 

PbCrO 4 1.8 X 10~ 14 (18C.) 

Ag 2 Cr0 4 9 X 10~ 12 (25C.) 

Fe(OH) 3 4 X 10- 38 

Mg(OH) 2 1.46 X 10- 11 (18C.) 

Bi 2 S 3 5.68 X 10~ 31 (18C.) 

BaS0 4 1 X 10- 10 (25C.) 

Agl 8.5 X 10-" 

solution of a difficultly or sparingly soluble electrolyte, the product of the 
concentration of the constituent ions is sensibly constant at any given 



124 QUALITATIVE ANALYSIS 

temperature. As can be seen from the latter example, if two of the 
ions produced are alike, the concentration of the ion is squared. In 
general, therefore, the concentration of each ion is raised to a power 
that is equal to the number of ions produced. Although this definition 
of the solubility-product principle seems ironclad, it is based upon an 
assumption, i.e., that the activity is equal to the concentration. Since 
in very dilute solutions, such as occur in the case of difficulty soluble 
substances, this is approximately true, we can use the solubility- 
product principle profitably. 

CALCULATION OF THE SOLUBILITY PRODUCT 

Example 1. Let us assume that we want to determine the K 8p for 
AgCl, the difficulty soluble silver compound that enables the chemist 
to isolate silver from a mixture. First of all, we should have to deter- 
mine how much silver chloride dissolves in 1 liter of water. This 
would be found to be 0.0015 gram per liter at room temperature (20C.). 
To transpose this quantity into molar concentrations, we divide 0.0015 

by the molecular weight of AgCl, namely, 143.34, and ^' Q , = 1.05 X 

14-O.O4 

10~ 6 . Since we assume that all the AgCl is dissociated into Ag + and 
Cl~, the concentration of Ag + would be 1.05 X 10~ 6 , and that of Cl~ 
would be 1.05 X 10~ 5 . Substituting these values in the equation 

K., = [Ag+][Cl-] 
and 

K 8P = (1.05 X 10~ 5 )(1.05 X 10-*) 
K ap = 1.1 X 10- 10 

Example 2. Magnesium hydroxide is soluble to the extent of 
0.0009 gram per 100 grams of water at 18C. What is the solubility 
product of this compound? Since the volume occupied by 0.0009 
gram of Mg(OH) 2 is small compared with the volume occupied by the 
100 grams of water, we may assume that the total volume is equal to 
the volume of the water, namely, 100 ml. Hence the molar concentra- 
tion of Mg(OH) 2 is equal to 

0.009 g./liter 
The molecular weight of Mg(OH) 2 = 58.33 

='0.000154 mole/liter of Mg(OH) 2 

The concentration of Mg++ = 0.0001 54M = 1.54 X 10~ 4 
The concentration of OH- = 2 X 1.54 X 10~ 4 = 3.08 X 10~ 4 
K.P = [Mg++][OH-] 2 = (1.54 X 10~ 4 )(3.08 X 10~ 4 ) 2 = 1.46 X 10~ 11 



REACTIONS IN SOLUTION 125 

Example 3. 0.000018 gram of Bi 2 S 3 dissolves in 100 grams of H 2 
at 18C. What is its K 9P 1 

0,000018 g./lOO g. o 0.00018 g./liter 

ft< S!?o g> = 0.00000035M = 3.5 X 10~ 7 

Ol4.^ 

C Bi +" = 2 X 3.5 X 10~ 7 = 7.0 X 10- 7 
C s - = 3 X 3.5 X 10~ 7 = 1.05 X lO' 6 
K 8P = [Bi+++] 2 [S~] 3 = (7.0 X 10~ 7 ) 2 (1.05 X 10~ 6 ) 3 
= (49.0 X 10- 14 )(1.16 X 10- 18 ) 
= 5.68 X 10~ 31 

Example 4. The K 8p of Ag 2 Cr0 4 at 25 is 9 X 10~ 12 . What is the 
molarity of this solution, and what is the solubility of Ag 2 Cr04? 
K. P = [Ag+] 2 [Cr()r] = 9X10 l2 

Let the number of moles of Ag 2 Cr0 4 present be represented by x. 
The concentration of Ag + = 2x. 

(2z) 2 (o;) = 9 X 10- 12 
4z 3 = 9 X 10- 12 
x 3 = 2.25 X 10~ 12 
x = \X2.25 X 10- 4 

x = 1.3 X 10~ 4 = molarity of Ag 2 Cr0 4 
Molecular weight of Ag 2 Cr0 4 = 332 

332 X 0.00013 = 0.043 g./litW = 0.0043 g./lOO ml. 

Application of the Solubility-product Principle. We now intend to 
show how this principle enables us to predict the course of a reaction, 
in which a difficultly or sparingly soluble precipitate appears as a reac- 
tion product. Let us suppose that we are interested in determining the 
[SO 4 M ] content of a solution. Could we precipitate the sulfate as 
BaS0 4 and then weigh it? To put it differently, does the reaction 

Ba++ + SOr ^ BaS0 4 

go sensibly to completion from left to right? In order to answer this 
question, we must know that the K sp for BaSO 4 is 1 X 10~ 10 . In 
a saturated aqueous solution of barium sulfate, the concentration of 
the constituent ions is, therefore, 1 X 10~ 5 gram mole per liter of solu- 
tion. We may compute from this the concentration of the respective 
substances in terms of grams per liter. 

[Ba++] = 1 X 10- 5 X 137.36 = 0.0013736 g./liter 
[SOr] ="1 X 10~ B X 96.07 = 0.0009607 g./liter 
[BaS0 4 ] = 1 X 10-' X 233.43 = 0.0023343 g./liter 



126 QUALITATIVE ANALYSIS 

Hence the solubility of barium sulf ate in aqueous solution is appreciable. 
If we precipitate from a volume of 250 ml. and use 150 ml. of water to 
wash the precipitate, we have used sufficient water to dissolve approxi- 
mately 1 mg. of barium sulfate. Of course, in actual practice, it is 
customary and necessary to use a slight excess of reagent (in our case, 
BaCU) in order to be certain that we have a stoichiometric quantity 
present and also to reduce the solubility of barium sulfate. If 250 ml. 
of solution has been employed, to which has been added sufficient 
barium chloride so that the final concentration of barium ions is 
0.001M (this requires only about 0.5 ml. excess of a 10 per cent solution 
of barium chloride), then 

[Ba-n-pCV] = 1 X 10- 10 

and 

i y in- 10 
[S0r]= \~~ 3 = 1 X 10-' 

and the concentration of sulfate ion remaining in solution will be 
10~ 7 X 96.07 = 0.000009607 g./liter 

and that of barium sulfate is 

10- 7 X 233.43 = 0.000023343 g./liter 

The solubility of bariun\ sulfate has been thus reduced to less than 
0.01 mg. in 250 ml. of solution. For the reason illustrated, it is 
invariably the practice in analytical chemistry to effect precipitations 
by adding a slight but distinct excess of reagent. 

CASE H. DILUTE ACETIC ACID AND SODIUM ACETATE; COMMON ION 
EFFECT AND BUFFER SOLUTIONS 

We shall start with the experimental fact that a dilute acetic acid 
solution is found to be weakly acidic. (A careful rereading of Chap. 
VII should make this apparently blatant truism somewhat less obvi- 
ous.) This can be explained by pointing out that the reaction between 
hydrogen acetate and water is incomplete and reaches equilibrium 
when the hydronium ion concentration is still relatively small. 

CHaCOOH + H 2 ^ CH 3 COO- + H 3 0+ 
Written in the mass action equation, the reaction becomes 

_ [CH 3 COO-][H30+] 
Ae " [CH 3 COOH] 

In a IM acetic acid solution the concentration of unchanged hydrogen 
acetate is 0.9958 moles. The concentration of the hydronium and 



REACTIONS IN SOLUTION 127 

acetate ions is therefore 0.0042 moles each. Substituting these values 
in the equation, we can calculate the equilibrium constant 

(0.0042) (0.0042) _ 

(0.9958) " '* * U 

It stands to reason that this constant should remain unchanged irre- 
spective of changes in the concentrations of the components. 

We now propose to perform a rather instructive experiment; let us 
drastically increase the acetate ion concentration by adding a sodium 
acetate solution. By using the equilibrium equation, we can predict 
that since K remains unchanged, increasing the concentration of one of 
the ions in the numerator should considerably decrease the concentra- 
tion of the other ion and also increase the denominator. To verify this 
contention experimentally, let us add 1 mole of sodium acetate to the 
foregoing acetic acid solution. Molar sodium acetate, according to 
Arrhenius, would have been declared to be 53 per cent ionized. Wo 
have become, however, more cautious about such things, and we shall 
merely admit that the effective available concentration of acetate ion is 
0.53 mole. Incorporating this value in the equation, we find that 

(0.0042 - x) (0.5342 - x) 



(0.9958 + x) 



= 1.8 X 



where x moles of acetate ion combine with x moles of hydronium ion to 
form x moles of acetic acid. Hence 

(0.5342 - x) = [CH 3 COO-] remaining 

(0.0042 - x) = [HsO + ] remaining 

(0.9958 + x) = [CH.COOH] after addition 

Expanding and solving for x, we get 

x = 0.004167 
or 

[H 3 0+] = 0.000033 mole/liter 
[CH 3 COO~] = 0.530033 
[CH 3 COOH] = 0.999467 

Thus the experiment has verified our prediction. But does it have any 
practical value? This can be answered by pointing out that it con- 
stitutes a procedure generally designated as "the common ion effect/' 
which is of utmost importance in analytical operations. To cite only 
two outstanding examples, the separation of Group II from Group III 
in cation analysis is based on this principle. (For a discussion of this 
case, see page 129). As our second example, we shall consider the 



128 QUALITATIVE ANALYSIS 

prevention of the precipitation of magnesium hydroxide in the third 
group. In this group, certain hydroxides, namely, those of iron, 
chromium, and aluminum, are prec pitated by the addition of ammo- 
nium hydroxide. If the solution contained magnesium ion, it would 
also be precipitated by the simple addition of ammonium hydroxide. 
To prevent this, we make use of the common ion effect. The solubility 
product of 

Fe(OH) 3 = 4 X 10~ 38 
Al(OH), = 1.9 X 10~ 33 
Cr(OH) 3 = 6.7 X 10~ 33 

whereas, on the other hand, the solubility product of magnesium 
hydroxide is 

Mg(OH) 2 = 1.46 X 10- 11 

Hence, if we could find a method of decreasing the hydroxyl ion con- 
centration, we could prevent the precipitation of magnesium hydroxide. 
If we add ammonium chloride, the increase in the ammonium ion con- 
centration will sufficiently lower the hydroxyl ion concentration and 
thus prevent the precipitation of magnesium hydroxide. 

Buffer Solutions. Let us refer again to the equilibrium equation of 
dilute acetic acid. 

_ [HsO+KCHaCOO-] 
A * ~ [CH,COOH] " X 

Assume now that we make the concentration of acetate ion equal to 
that of hydrogen acetate. Then 

- 1.8 x 10- 



It can be seen that the concentration of the hydronium ion has become 
equal to that of the equilibrium constant 1.8 X 10~ 6 . The irnplica- 
ions are quite apparent. In numerous chemical and, more par- 
ticularly, biochemical processes, it is imperative to maintain a definite 
hydronium ion concentration fixed within narrow limits. Numerous 
microorganisms, for instance, refuse to grow unless the [H 3 + ] is 
adjusted to a certain level. Solutions of the sodium acetate-hydro- 
gen acetate type are therefore widely used. They are called buffer 
solutions and are prepared by selecting appropriate concentrations of 
weak acids in equilibrium with the corresponding anions. Two hun- 
dred milliliters of a solution containing 50 ml. of 0.2M primary 
potassium phthalate and 0.4 ml. of 0.2M NaOH will maintain a hydro- 
nium ion concentration of 10~ 4 or a pH of 4. 



REACTIONS IN SOLUTION 129 

The name buffer is quite appropriately chosen. If we refer once 
more to the acetic acid equilibrium, it can readily be shown that the 
presence of a larger amount of sodium acetate would resist and mini- 
mize the accidental introduction of some strong acid, and that the 
reaction 

H 3 0+ + (C 2 H 3 2 )- -> HC 2 H 3 2 

would take place and thus maintain the original H 3 + ion concentra- 
tion. This mechanism is vital to us in the most literal sense of the 
word. Human blood must maintain a H 3 O + concentration correspond- 
ing to a pH of 7.6. Any deviation from this level would lead to serious, 
perhaps fatal, injury. Blood is therefore protected by the presence of 
such buffers as bicarbonates, phosphates, etc. 

CASE III. FORMATION OF SLIGHTLY IONIZED PRODUCT 

When we deal with difficultly soluble salts of weak acids, it is possi- 
ble to regulate the hydronium ion concentration of the solution so as to 
control with very high precision the concentratioh of the anions of the 
weak acids. This is, in brief, tho principle upon which is based the 
separation of the metal sulfides of the hydrogen sulfide group from 
the ammonium sulfide group in the ordinary qualitative scheme. 

Separation of the Cations of Group II from Those of Group III. By 
purely empirical methods, the pioneers of analytical chemistry had 
found that, whenever a solution containing the ions of Groups II and 
III was made Q.3N with hydrochloric acid and saturated with hydrogen 
sulfide, only the sulfides of Group II, i.e., cupric sulfido, bismuth 
sulfide, arsenic sulfide, etc., would precipitate. The other ions present 
were left in solution. This method of separation has proved to be so 
convenient that it is retained in practically all the analytical schemes. 
Careful experiments have shown that this separation is definitely 
dependent upon the hydronium ion concentration. If the analyst 
should become careless and try to precipitate the sulfides from a solu- 
tion of greater acidity (e.g., one that is 37V), he would find that a num- 
ber of the cations of the second group, e.g., Cd ++ , Sn++, Pb+ + , Bi+++, 
would not form any insoluble sulfides. On the other hand, a solution 
only 0.027V with respect to the hydronium ion would allow the precipi- 
tation of zinc sulfide, which is a member of Group III. 

Let us now show how all these experimental facts can be accurately 
predicted with the help of the common ion principle and a chemical 
handbook. The reaction of H 2 S with water can be represented by the 
following equations: 

H 2 S + H 2 ^ H 3 0+ + HS- 
HS- + H 2 ^ H 3 0+ + S- 



130 QUALITATIVE ANALYSIS 

Hence the equilibrium constant for these reactions would be formulated 
thus: 

jp- ^HaO* X C/HS- 

*M ~ 7y 

OH Z S 
and 

v CHSO+ X Cs- 



Multiplying K\ by K 2 , we get 

//"Y \2 \/ ft 

K = K K = ^ H3 + / * ^ s 



At room temperature (25C.) and 1 atm. pressure, a liter of water 
saturated with hydrogen sulfide contains 0.1 mole of H 2 S. 
In this solution the 

H 3 0+ concentration = 9.5 X 10~ 5 
HS- concentration = 9.5 X 10~ 6 

S- concentration = 1.2 X 10~ 15 

Substituting these values in the expressions for the equilibrium con- 

stants, we obtain 

Ki = (9.5 X 10-*K9.5 X 10-Q m 9 x 1Q _ 8 

_ (9.5 X 10- 6 )'(1.2 X 10-") _ j 
Az (9.5 X 10- 8 ) ~ 1<J X 1U 



= (9 X 10-)(1.2 X 10- 18 ) = 1.1 X 10- 22 

We have thus calculated the equilibrium constant for a saturated 
solution of hydrogen sulfide. We can now write the equilibrium 
expression in this form: 

(cw>+) 2 x c s - = (i.i x io- 22 )(o.i) = 1.1 x 10-23 

and 

_ (i.i x io- 23 ) 



_ 



This equation shows obviously what the relation is between the 
hydronium ion concentration and the sulfide ion concentration. The 
sulfide ion concentration is inversely proportional to the square of 
the hydronium ion concentration. A tenfold increase in the 
hydronium ion concentration decreases the sulfide ion concentration a 
hundredfold. 

Let us now show how some simple calculations would permit us to 
predict that, if a solution containing cadmium ion and zinc ion is made 



REACTIONS IN SOLUTION 131 

0.3AT with hydrochloric acid and then saturated with hydrogen sulfide, 
only cadmium would precipitate as the sulfide. Let us use the equa- 
tion obtained above and substitute 0.3 for the hydronium ion concen- 
tration (the amount of HsO 4 " formed by the dissociation of the H^S.is so 
small that it may be neglected) and thus calculate the concentration of 
the sulfide ion in this solution. 

c = (u x 1Q ' 23) = (L1 x io- 23 ) 

s " (0.3) 2 0.09 

C s - = 1.2 X 10~ 22 

Let us assume that the solution that we wish to test is 0.0 IM with 
respect to cadmium ion and 0.0 IM with respect to zinc ion. The solu- 
bility products of the corresponding sulfides are 

CdS, K 8P = 3.6 X IO- 29 
ZnS, K 8P = 1.2 X IO- 23 

Substituting the values obtained previously, we have 

ZnS, K = [Zn++][S~] - (l.l X 10- 22 )(0.01) 
K = 1.1 X IO- 24 

Sincd K sp > K, zinc sulfide would not precipitate; i.e., the product of 
the sulfide ion concentration and the zinc ion concentration is smaller 
than the value of solubility product. On the other hand, for cadmium 
sulfide, we have 

K = [Cd++][S-] = (1.1 X 10~ 22 )(0.01) 
K = 1.1 X IO- 24 

Since here K > K sp , cadmium sulfide would precipitate. 

We may also compute the concentration of the cadmium ion remain- 
ing in solution after the reaction has reached equilibrium. 

[Cd ++ ][S-] = 3.6 X IO- 29 
[Cd++](l.l X IO- 22 ) = 3.6 X IO- 29 

Q A V 1 0~ 29 

fi = 3.3 X 10-7 



[Cd++] = 0.00000033 mole/liter 

This fact, expressed in words, plainly indicates that the amount of 
cadmium ion remaining in solution is negligible. We have selected one 
representative ion in both the hydrogen sulfide group and the ammo- 
nium sulfide group. We should like to suggest to the student who 
possesses both a healthy curiosity and an appropriate amount of 
skepticism that he test the general principle illustrated above by 
calculating the ion products for the other members of Group II and of 



132 QUALITATIVE ANALYSIS 

Group III and compare them with the solubility products of the cor- 
responding sulfides. 

CASE IV. FORMATION OF A VOLATILE GAS 

Let us assume that a solution containing Ba + + has been treated 
with (NH 4 ) 2 CO 3 solution, thus forming a precipitate of white, powdery 
BaC0 8 . This is a procedure commonly resorted to in qualitative 
analysis to isolate the alkaline-earth group. Barium carbonate is the 
salt of a moderately strong base and a weak acid. Its solubility prod- 
uct could be written as 

[Ba++][CO,-] = 8.1 X 10- 9 
from the equation 

^ Ba++ + C0 3 



If, to a saturated aqueous solution of barium carbonate in contact with 
solid barium carbonate, we add a small amount of a strong acid (let us 
say hydrochloric acid), there will be a reaction with the carbonate ions, 
resulting in the formation of carbonic acid, which will further break 
up into C0 2 and H 2 0. 

COr + H 3 0+ - HCCV + H 2 
HCO 3 - + H 3 0+ -> H 2 C0 3 - H 2 + C0 2 t 

In this way, the concentration of the carbonate ion will be reduced, and 
the equilibrium will be displaced to the right. In other words, barium 
carbonate will dissolve in an attempt to maintain an ion concentration 
sufficient to satisfy the solubility product. Under the same circum- 
stances, barium sulfate would not dissolve, because the presence of a 
nonvolatile acid would not materially affect the concentration of the 
sulfate ion. In the same way, a salt of a volatile base and a strong acid 
may be dissolved by the addition of a solution of a strong base to its 
saturated solution in contact with the solid salt. 

CASE V. FORMATION OF A COMPLEX ION 

The fact that silver chloride is soluble in a solution of ammonia is 
applied to the separation of the metals of the first group. It is cited 
here to illustrate the relationship between solubility product and com- 
plex ion formation. If we have a saturated solution of silver chloride 
in contact with solid silver chloride and add ammonia solution, the 
silver chloride will dissolve if enough ammonia has been added. 

AgCl ^ Ag+ + Cl- 



REACTIONS IN SOLUTION 133 

However, the diamminosilver ion dissociates to a certain extent: 

Ag(NH 3 ) 2 + ^ Ag+ + 2NH 3 
The equilibrium constant for this reaction is 

[A*1[NH,] R 
K ~ - 6 ' 8 X 



In order to examine the effect of an ammonia solution on the equilib- 
rium between Ag+ and Cl~ and solid AgCl, let us assume the following 
conditions: a solution of silver nitrate whose total [Ag+] is 0.01M has 
been treated with NH 3 solution until the concentration of the latter is 
O.lAf. We now ask the question, what is the concentration of free 
[Ag+] remaining in this solution? Utilizing the equilibrium expression 

[Ag+][NH 3 P 

- 6 - 8 x 10 



we can readily find the answer if we incorporate the following known 
concentrations into the formula. Since the original silver ion con- 
centration is 0.01M and since Ag(NH 3 ) 2 + is only slightly dissociated, 
we may assume that 

[Ag(NH 3 ) 2 +-] = 0.01M 

The [NH 3 ] is 0.1 ; therefore [NII 3 ] 2 - 0.01. Incorporating these values 
in the equilibrium expression, we have 

[Ag+]X(0.01) _ 68ylo - 8 

~~ " b ** x IU 



from which follows that the [Ag+] = 6.8 X 10~ 8 . In other words, 
under the stipulated conditions and after the formation of the silver 
ammonia complex, 6.8 X 10~ 8 mole of Ag+ remains in solution. Let us 
now investigate this system further in order to determine the chloride 
ion concentration required to cause a precipitate of AgCl. The solubil- 
ity product of AgCl is 1.1 X 10" 10 . Let us now calculate the [C1-] that 
must be exceeded in order to cause precipitation. 

K 8P = [Cl-][Ag+] = 1.1 X 10-' 
[Cl-](6.8 X 10- 8 ) = 1.1 X 10- 10 

1 1 V 10~ 10 

t ch l - - L62 x 



In the absence of ammonia solution and the same total concentration of 
silver ion as silver nitrate (O.OlAf), precipitation would have resulted 
when the chloride ion concentration exceeded 1.1 X 10~ 8 . 



134 QUALITATIVE ANALYSIS 

One interesting observation is that whereas silver chloride is soluble 
in an ammonia solution, silver iodide is insoluble in ammonia. This 
fact can be explained on the basis of the solubility product of silver 
iodide. 

K 8P = 8.5 X 10- 17 

That is, a saturated solution of silver iodide contains the following 
silver ion concentration: 

[Ag+][I-] = 8.5 X 10- 17 
since [Ag+] = [I~] 

[Ag+] 2 = 85 X 10- 18 
[Ag+] = 9.2 X 10- 9 

But the concentration of silver ion furnished by the dissociation of the 
diamminosilver complex is 6.8 X 10~ 8 . This means that since the 
dissociation of the diamminosilver ion furnishes more silver than is 
necessary for silver iodide to precipitate, adding iodide ion to q, solution 
of silver chloride in ammonia would result in the precipitation of silver 
iodide. 

K, the constant that governs the equilibrium between the complex 
ion and its components, is customarily designated as the instability 
constant. The j ustificatioii for this term can be readily realized when 
we consider that the smaller the instability constant the smaller the 
concentration of free ions in equilibrium with the complex ion. 

Another important application of this principle is found in the 
separation of cyanides of copper and cadmium by means of hydrogen 
sulfide. The equilibrium constants for the dissociation of these com- 
plex cyanides are 

Cu(CN) 3 ~ ^ Cu+ + 3CN- 
' _ 

" & * IU 



Cd(CN)r ;= Cd++ + 4CN~ 
_ [Cd+*][CN-' 



[Cd(CN)r] 



- 1-4 X 10 



We can see that the instability constant for cadmium is considerably 
greater than that for copper. When hydrogen sulfide is introduced, 
the sulfide ion concentration is great enough to exceed the solubility 
product of the cadmium sulfide but not great enough to exceed that of 
the cuprous sulfide. Hence this is a commonly used separation 
procedure. 



REACTIONS IN SOLUTION 135 

READING REFERENCES 

GETMAN, F. H., and F. DANIELS: "Outlines of Theoretical Chemistry," 6th ed., 

John Wiley & Sons, Inc., New York, 1937. 
GLASSTONE, S.: " Textbook of Physical Chemistry," D. Van Nostrand Company, 

Inc., New York, 1940. 
HAMMETT, L. P.: "Solutions of Electrolytes," 2d ed., McGraw-Hill Book Company, 

Inc., New York, 1936. 

QUESTIONS 

1. The solubility of AgCl at 100C. is 0.002 g./lOO g. of water. Calculate the 
solubility-product constant. 

2. The solubility of lIg 2 Cl 2 at 40C. is 0.0007 g./lOO g. of water. Compute the 
solubility-product constant. 

3. At 20C., the solubility of Ag 3 Fe(CN) 6 is 6.6 X 10 6 g./lOO g. of water. 
Calculate the solubility-product constant. 

4. The solubility product of CaCO 3 is 0.87 X H)- 11 ,-it 25C. Calculate the 
molarity of the solution and the solubility of CaCO 3 . 

6. If we have 100 ml. of an O.OOU/ solution of AgNO 3 and if 100 ml. of O.OOUf 
HC1 is added, compute the concentration of the Ag + that remains unprecipitated. 
If 0.1 ml. of the HC1 is used in excess, what would be the concentration of Ag + , 
which remains unprecipitated? 

6. What concentration of I1C1 is necessary to prevent the precipitation of CdS 
by S-? 

7. What is the hydronium ion concentration of I 1. of an 0.05717 solution of 
acetic acid to which has been added 100 ml. of 0.1 M solution of sodium acetate? 
Calculate the pFI. 

8. How many ml. of a 2M solution of sodium acetate must bo added to a liter 
of an 0.02M solution of acetic acid in order to make the pH of the solution 7.0? 

9. Using the data from Prob. 4, calculate the concentration of Ca ++ that must 
be added to a solution of CO 2 at 30C. and 1 atm. pressure in order just to form a 
precipitate? (Solubility of CO 2 at 30C., 0.1257 g./lOO g. of water.) 



CHAPTER IX 

REACTIONS OF CHEMICAL COMPOUNDS 
IV. OXIDATION-REDUCTION REACTIONS 

To predict the outcome of reactions or even to predict the dis- 
covery of facts as yet unknown is more or less routine for the scientist; 
even a third-rate scientist must be fairly proficient in this art. We 
shall now consider some rarer achievements that demand either genius 
or considerable luck and sometimes both. When Louis Pasteur 
showed that the lowly yeast cells were responsible for the fermentation 
of grape juice, he not only explained the making of wine but, more 
significantly, correlated two sets of facts that had seemed until then to 
be unrelated. Such a correlation, taken from a different field, will 
provide the subject matter of this chapter. We propose to study this 
question: Are there any direct relationships among phenomena, such as 
the disappearance of the permanganate color upon the addition of 
ferrous ions and the electric current that flows through the wires in our 
streets and homes and that dispels the darkness and runs our vacuum 
cleaners? 

The Experimental Basis. In order to answer this question, we 
must first describe the peculiar type of chemical reaction that we wish 
to study. All students of qualitative analysis have at one time or 
another performed the test for the stannous ion with mercuric chloride 
solution. The presence of the stannous ion is indicated by the forma- 
tion of a white precipitate that slowly turns gray and then black. 
The reactions that occur can be represented by the following equations : 



Sn++ -> 
2C1- + 2Hg++ -* Hg 2 Cl 2 - 2Hg + 2C1~ 

Essentially, the reactions amount to this: the stannous ion has gained 
positive charges, and the mercuric ion has lost positive charges. We 
are accustomed to calling the gain of positive charges (or increase in the 
valence number) oxidation 1 and the loss of positive charges (or decrease 

1 The term oxidation, with its suggestion of oxygen, has been retained out of 
respect for the pioneering work of Lavoisier, who thought that oxidation invariably 
involves union with oxygen. At present, we use the term oxidation to characterize 
a much wider range of phenomena. 

136 



OXIDATION-REDUCTION REACTIONS 137 

in valence number) reduction. Thus far, we have only a definition, no 
attempt having been made to determine how this change in the positive 
charge has been brought about. Let us remember, however, that 
no oxidation is possible if a reduction does not occur simultaneously. It 
is very much like a transaction on the Stock Exchange; the gain of Mr. 
X must be accompanied by a loss on the part of Mr. Y. 

Let us illustrate this point with another example. When gaseous 
chlorine is passed into a solution containing ferrous ions, we find that 
the following changes take place: 

Fe++ - Fe+ ++ 
C1 2 - 2C1- 

Again, the oxidation of the ferrous ion has proceeded simultaneously 
with the reduction of the chlorine. Note that the substance that is 
being reduced, the chlorine, can be called an oxidizing agent, since it is 
the means by which ferrous ion is being oxidized, and similarly, the 
ferrous ion that is being oxidized can be called a reducing agent. 

The Electronic Explanation. The next question to be answered is, 
what happens structurally to an atom when it gains an additional 
positive charge? We already know the answer to this question from 
a previous discussion. Let us recall the transformation of the 
sodium atom into the sodium ion. The electrically neutral sodium 
atom, having 11 positive charges on the nucleus neutralized by 11 
electrons in its electronic orbits, lost 1 electron, thus leaving 1 1 positive 
charges and 10 electrons. The net effect was that the sodium ion, 
thus produced, had a charge of +1. 

Na - Ic - Na+ 

Using the same kind of electronic bookkeeping for Fe 44 *, we should say 
that the loss of one electron leads to 

Fe++ - le - 

We can now define oxidation as the loss of electrons and reduction as 
the gain of electrons. In oxidation-reductions systems, some ions or 
atoms gain electrons, which are, in turn, lost by other ions or atoms; 
however, the whole system remains electrically neutral, as it was from 
the start. 

THE CORRELATION 

Now an interesting point arises; oxidation and reduction seem to 
involve the transfer of electrons from one substance to another. We 
know of another case wherein the transfer of electrons occurs, namely, 
in the passage of the electric current through a wire. Now, then, is 



138 



QUALITATIVE ANALYSIS 



there any correlation between these two phenomena? Is it necessary 
that the electrons be furnished by another substance in contact with it, 
or can we use an electric current to supply the electrons? 

The Experimental Evidence. The easiest way and the scientific 
way to find out is to make an experiment. Into each of two beakers A 
and B, A containing a solution of ferrous chloride and B containing a 
solution of ferric chloride, let us immerse a wire. Next, we shall con- 
nect the wire from beaker B to the negative pole of a battery and that 




FIG. 38. Oxidation and reduction by an electric current. 



from beaker A to the positive pole of the battery. To complete the 
circuit, let us now dip a finger from one hand into the solution in beaker 
A and a finger from the other hand into the solution in beaker B. 
After a little while, if we test for the presence of ferric ion in beaker A, 
we shall find it present, and similarly, we shall find ferrous ion in 
beaker B. Here, indeed, we have proved that ferrous ion can be con- 
verted to ferric ion by the removal of electrons and vice versa. Evi- 
dently, then, oxidation-reduction is an electrical phenomenon and 
can be brought about by an electric current through the transfer 
of electrons. 

However, we know that if we put ferrous and ferric ions together 
in one beaker, nothing will apparently happen. However, if we mix 
other substances, such as stannous chloride and mercuric chloride, there 
will be spontaneous oxidation and reduction. The question then 
presents itself, does the stannous chloride, of necessity, have to be in 
contact with the mercuric chloride, i.e., be in the same beaker? To 
answer this question, let us again perform an experiment. Let us take 
two beakers, fill one with stannous chloride solution and the other with 



OXIDATION-REDUCTION REACTIONS 



139 



mercuric chloride solution, dip a wire into each, connect the wires to a 
galvanometer, and dip a finger into each beaker to complete the circuit. 
The galvanometer needle will be deflected, showing the passage of 
current, and mercury and mercurous chloride will be in one of the 
beakers, showing that the oxidation and reduction are proceeding. 
Now we have proved the point that we wished to prove, namely, that 
a spontaneous oxidation-reduction actually means that electrons are 
being transferred and that the transfer can be from ion to ion or through 
an intervening external circuit. 




FIG. 39. Production of a flow of electrons by a spontaneous oxidation-reduction. 

The Salt Bridge. One question still remains obscure. The inquir- 
ing* student may ask, if the oxidation-reduction proceeds through the 
transfer of electrons, why do we need to complete the circuit by insert- 
ing a finger into each beaker? The answer is not too difficult to find. 
In the first experiment, in the beaker containing the ferric chloride, we 
had three chloride ions for every ferric ion, and hence the solution was 
electrically neutral. If we reduced some of the ferric ion to ferrous ion, 
there would have been an excess of chloride ion, since we need but two 
chloride ions for every ferrous ion ; hence the solution would have been 
negatively charged. This charge would have opposed the further flow 
of electrons into the solution; hence we provided an exit for the excess 
chloride ions through our body. Similarly, the other beaker would 
have been lacking in chloride ion, and we therefore supplied them 
through our body. 

A better practice than using the human body as a return circuit 
would have been to connect each beaker to a reservoir containing a 
large number of ions. For instance, we could have connected each 



140 



QUALITATIVE ANALYSIS 



beaker to the ocean by means of a wet rag containing an electrolyte. 
In this case, the beakers could have been separated by miles of dis- 
tance, with only a wire connecting them. 




New York 



Wet rcrg 



-zr Ocecrn Ocean 

FIG. 40. Oxidation and reduction using the ocean as a reservoir for ions. 

A still better method, which is generally used in laboratories, is to 
connect the two solutions by a salt bridge. A salt bridge is merely a 

glass tube, filled with a saturated 
solution of an electrolyte (usually 
KC1), which is closed by a glass 
stopcock, or a glass plug. This 
obstruction prevents gross diffu- 
sion of the solution but does not 
prevent the migration of the ex- 
cess ions. 

Formal Definitions. As aeon- 
elusion, we should again state 
formally the definitions of oxlda- 



Closed stopcock 




FIG. 41. Oxidation. and reduction using a 
salt bridge. 



tion and reduction. Oxidation is 
the loss of electrons by an atom or ion, or the algebraic increase in 
the valence number of the element. Reduction is the gain of electrons 
by an atom or ion, or the algebraic decrease in the valence number of 
the element. 

BALANCING OXIDATION-REDUCTION EQUATIONS 

The simple oxidation-reduction equations can usually be balanced 
by trial. However, when more complicated equations are encoun- 
tered, the simpler methods are extremely tedious and uncertain and, 
indeed, often give us equations that, although they are balanced in the 
sense that the same number of atoms of each element appear on each 
side of the equation are, nevertheless, incorrectly balanced, since they 
do not represent what actually happens. However, by using the 
method described in the following pages, one can balance all oxidation- 



OXIDATION-REDUCTION REACTIONS 141 

reduction reactions with dispatch and with the certainty that they will 
be correct stoichiomctrically. 

Example 1. Let us first take a very simple case that can be rapidly 
balanced by inspection and see how the method works. 

Skeleton equation: 

SnCl 2 + HgCl 2 -> Hg 2 Cl 2 + SnCl 4 (1) 

Let us first write this ionically : 
Skeleton equation: 

Sn++ + 2C1- + Hg++ + 2C1- -> Hg 2 Cl 2 + Sn++++ + 4C1~ (2) 

Next let us determine which ions have been reduced and which ones 
oxidized. This can often be done by inspection or, in more compli- 
cated cases, by calculating the valence numbers of each element and 
then finding by inspection which valence numbers have increased 
and which ones have decreased. In this case, tin has gone from a 
valence number of +2 in Sn+ + to +4 in Sn ++++ ; hence it has been 
oxidized. Writing the equation for this partial or half reaction, we get 

Sn++ - 2e -> Sn++++ (3) 

We must subtract two electrons from the left-hand side of the equation 
in order to make both sides equal in electrical charge. The mercuric 
ion has been reduced, going from +2 in Hg ++ to +1 in Hg 2 Cl 2 . Again 
writing the half equation, we get 

2C1- + 2Hg++ + 26 -> Hg 2 Cl 2 (4) 

Adding equations (3) and (4), we get 

2C1- + 2Hg++ + Sn++ + 2e - 2e -> Hg 2 Cl 2 + Sn++++ (5) 



The electrons that we have added and subtracted cancel, and we get 
2C1- + 2Hg++ + Sn++ -> Hg 2 Cl 2 + Sn++++ (6) 

Equation (6) tells us that every stannous ion reduces two mercuric ions 
to one mercurous chloride molecule. This equation (6), as written, 
has, however, a +4 charge on each side of the equation. To write a 
reaction between neutral salts, we- must add four chloride ions to each 
side and so render each side neutral, as 



6C1- + 2Hg++ + Sn++ - Hg 2 Cl 2 + Sn++++ + 4C1- (7) 

Example 2. Let us now take a more complicated case and go 
through the same steps. 



142 QUALITATIVE ANALYSIS 

Skeleton equation: 

KMn0 4 + FeSO 4 + H 2 SO 4 - MnSO 4 + Fe 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O 

(8) 

Written ionically 
Skeleton equation: 

K+ + Mn0 4 ~ + Fe++ + SOr + H 3 + -> 

Mn++ + SOr + FC+++ + K+ + H 2 O (9) 
Valence number of manganese in Mn0 4 ~ = +7 

in Mn++ = +2 

The permanganate ion has been reduced. 

Valence number of iron in Fc ++ = +2 
in FC+++ = +3 

The ferric ion has been oxidized. 
The half equations are 

8H 3 0+ + MnOr + 5e -> Mn++ + 12H,0 (10) 

Fe++ - le->Fe+++ (11) 

NOTE: To balance equation (10), we added hydro nium ion, which we com- 
bined with the oxygen of the permanganate ion to form water. This* may not be 
the true mechanism, but it is conventionally agreed that we shall do this except 
where the equation indicates that it is impossible. 

Equations (10) and (11) are not, however, electrically equal, and if we 
added them as such, the electrons that we have added and subtracted 
would not cancel; hence we multiply equation (11) by 5 and then add 
the two equations. 

8H 3 O+ + MnO 4 ~ + 5e -> Mn++ + 12H 2 O 
5Fe + + - 5c 



____ __ 

8H 3 6+ + MnOr + 5Fe++ -> Mn++ + 12H 2 O + 5Fe+++ (12) 

This is the equation for the oxidation-reduction. The complete 
equation may now be written by inspection. 

K+ + MnO 4 - + 5Fe++ + 8H 3 O + 9S0 4 - - 

Mn ++ + 9SOr + 5Fe ++ + + 12H 2 + K+ (13) 

Example 3. 
Skeleton equations: 

CrCl 3 + Mn0 2 + H 2 ~> MnCl 2 + H 2 Cr0 4 (14) 

Cl- + Mn0 2 + H 2 - Mn^ + + Cl~ + H 3 0+ + Cr0 4 - 

(15) 



OXIDATION-REDUCTION REACTIONS 143 

Half equations: 



12H 2 + Cr+++ - 3 
4H 3 0+ + Mn0 2 + 


t - CrO 4 - + 8H S 0+ 2 
2e -* Mn++ + 6H 2 3 


(16) 
(17) 

(18) 
(19) 


24H 2 + 2Cr+++ - 
12H 3 0+ + 3Mn0 2 -f 


6 - 2Cr0 4 = + 16H 3 O+ 
- 6e - 3Mn++ + 18H 2 () 


6H 2 + 2O+++ + 3Mn() 2 -> 2Cr0 4 ~ + 4H 3 O+ + 3Mn++ (20) 
6H 2 + 2O+++ + 6C1- + 3MnO 2 - 
4H 3 0+ + 2CrOr + 3Mn++ + 6C1~ (21) 


Example 4. 






Skeleton equations : 

KMn0 4 + MnSO 4 
K+ + MnO 4 ~ + Mn++ + 


+ H 2 - K 2 SO 4 + Mn(), + H 2 S0 4 (22) 
SO 4 - + H 2 - 
K+ + SOr + MnO + H 3 0+ (23) 


It is evident here that the permanganate ion is being reduced to MnOa 
and that the manganous ion is being oxidized to MnO 2 . 


Half equations: 






MnO 4 - + 4H 3 0+ + 3e -> MnO 2 + 6H 2 < 
Mn++ + 6H 2 - 2 -> Mn0 2 + 4H,<)+ 


1 2 (24) 
3 (25) 


2MnO 4 - + 3Mn ++ + 6H 2 O ~> 4H 3 O + + 5Mn() 2 (26) 
2K+ + 2Mn0 4 ~ + 3Mn++ + 6H 2 () + 3SO 4 - -> 
4H 8 0+ + 3SOr + 2K+ + 5Mn0 2 (27) 


Example 5. 






Skeleton equations: 






Hg + HNO, -> Hg(NO 3 ) 2 + II 2 
Hg + H 3 0+ + N0 3 ~ ~> Hg++ + NOr + 


+ NO (28) 
H 2 + NO (29) 


Half equations: 






Hg- 
4H 3 0+ + NO 3 - 4 


2 - Hg++ 3 
- 3 - NO + 6H 2 2 


(30) 
(31) 


3Hg + 8H 3 O+ + 
3Hg + 8H 3 0+ + 8NOr 


2NO 3 ~ ~> 3Hg++ + 2NO + 12H 2 O (32) 
- 3Hg++ + 6NOr + 2NO + 12H 2 (33) 



Example 6. 
Skeleton equations: 

AuCl 3 + H 2 O 2 + NaOH - NaCl + H 2 O + O 2 + Au (34) 
AU+++ + Cl- + H 8 0+ + Or + Na+ + OH- - 

Na+ + Cl- + H 2 + 2 + Au (35) 



144 QUALITATIVE ANALYSIS 

Half equations: 

AU+++ + 3e Au 



(36) 

(37) 

2Au- t ~ H - + 30 2 " - 2Au + 30 2 (38) 

2Au+++ + 6C1- + 6Na+ + 6H 3 0+ + 60H- + 3O 2 - - 

2Au + 30 2 + 12H 2 O + 6Na+ + 6C1- (39) 

Example 7. 
Skeleton equations: 

CoCl 2 + KOH + H 2 O 2 -> KC1 + Co(OH) 3 (40) 
Co++ + Cl- + K+ + OH- + H 3 O+ + 0,- - 

K+ + Cl- + Co(OH) 3 + H 2 O (41) 

Half equations: 



30H- + Co++ - e - Co(OH) 3 
O 2 " + 2H 2 O + 2 -4 4OH- 



(42) 

(43) 

2OH- + 2Co++ + Or + 2H 2 O -> 2Co(OH) 3 (44) 

2Co++ + 4C1- + 4K+ + 40H- + 2H 3 0+ + Or + 2H 2 -* 

2Co(OH) 3 + 4K+ + 4C1- + 4H 2 O (45) 

THE MEASUREMENT OF THE OXIDIZING POWER 

One other problem is yet to be considered, namely, if we wish to 
oxidize a substance in the laboratory, can we choose any known oxidiz- 
ing agent? Upon consideration, the answer will be, no, we cannot 
choose any oxidizing agent. For although it is true that the primary 
purpose of an oxidizing agent is to gain electrons and so be reduced, we 
must remember that we are dealing here, as always, with opposing 
tendencies the tendency of an ion to gain electrons or to lose them. 
If we mix two substances that can either gain or lose electrons, the one 
that has the greater attraction for electrons will gain them at the 
expense of the other. If now we take the substance that in this first 
reaction has been shown to be able to lose electrons and put it in the 
presence of a substance that loses electrons still more readily, it will 
gain them. A good example of an ion that does this readily is the 
nitrite ion. In the presence of the iodide ion, the nitrite ion acts as an 
oxidizing agent. 

21- - 2e -> I 2 
2H 3 0+ + N0 2 - + c - NO + 3H 2 

But in the presence of the permanganate ion, which is a stronger 
oxidizing agent, the nitrite ion acts as a reducing agent. 



OXIDATION-REDUCTION REACTIONS 145 

8H 3 0+ + MnOr + 56 -> Mn++ + 12H 2 
3H 2 + N0 2 - - 2* -> N0 3 - + 2H 3 0+ 

Accordingly, we are interested in the question, how strong is an oxidiz- 
ing agent, or how great an affinity does it have for electrons? In 
ordefr to answer this question adequately, we must investigate the 
devices that permit the measurement of the oxidizing power of a 
substance. 

Since the oxidizing power is concerned with the intensity of the 
desire of the ions to gain electrons and since an electron flow is an 
electric current, measuring the intensity of the electric current, i.e., its 
potential, will provide us with an answer. Suppose that we have in a 
beaker a solution of a strong oxidizing substance, such as eerie sulfate. 
Ceric ions, Ce ++++ , tend to acquire electrons giving cerous ions, as 

Ce 4 " f + + + c - Ce++ + 

When we begin our experiment, the oxidized form Ce ++4 " f is over- 
whelmingly preponderant, and the concentration of the reduced form 
Ce ++ + is practically negligible. On introducing a platinum electrode 
into this solution, what will happen? Ceric ions will remove some 
electrons from the platinum electrode, and the electrode will acquire a 
positive charge. Now we wish to measure the magnitude of this 
charge. We must remember, however, that this electrical potential, 
the intensity factor of electrical energy, can be measured only if it is 
compared with another known potential. We therefore connect the 
eerie half cell with another beaker containing a reference electrode, i.'e., 
an electrode that maintains a constant charge. Suppose we now find 
that the difference between the potentials of the platinum electrode 
in the eerie solution and the reference electrode is + 1 .70 volts. We can 
now say, let us take the potential of the reference electrode to be zero; 
hence the potential of this eerie solution is +1.70 volts. 

This assumption is a defensible one. The same assumption is made in measur- 
ing the intensity factor of heat energy, the temperature, where the melting point 
of ice is taken to be zero and all other temperatures are measured relative to it. 

Reference Electrodes. In our experiment, we used an electrode 
that would maintain an unvarying potential. The electrode that has 
been chosen as the reference electrode and hence has been assigned a 
potential of zero is the hydrogen electrode. It consists of a platinum 
electrode, over which flows gaseous hydrogen at 1 atm. pressure. The 
platinum is coated with platinum black, which is finely divided platinum 
metal. This platinum black serves as a catalyst, accelerating the 
attainment of equilibrium. The electrode is in contact with an acidic 



146 QUALITATIVE ANALYSIS 

solution having a hydronium ion activity of 1 (i.e., [H 3 + ] of 1 approxi- 
mately). Let us now see why a constant potential is maintained under 
these conditions by this half cell. The hydrogen molecules absorbed 
on the electrode lose electrons giving hydronium ions to the solution 
thus: 

2H 2 + H 2 - 2c - 2H 3 0+ 

The electrode thus acquires a negative charge, but the reverse reaction 
also occurs, namely, hydronium ions gain electrons from the electrode 
and form hydrogen, as 

2H 3 O+ + 2 - H 2 + 2H 2 O 

This tends to impart a positive charge to the electrode. After equilib- 
rium has been attained, the two reactions will impart to the electrode 
a definite potential, which will remain constant as long as the concen- 
trations of the 112 and the II 3 + remain unchanged. 

If need be, we can, by the use of such a reference electrode, calibrate 
other .reference electrodes, whose potential we shall then know. One of 
the most common auxiliary reference electrodes used is the calomel 
electrode. It consists of mercury covered by a paste of mercurous 
chloride, potassium chloride, and mercury in contact with a solution 
saturated with mercurous chloride and potassium chloride. Contact is 
made with the mercury by means of a platinum wire. 

Molar Potentials. Now let us return to our example, in which a 
solution containing cerous and eerie ions in contact with a platinum 
electrode is connected to our standard reference electrode. In our 
example, the eerie ions were preponderant, and the reduced form, the 
cerous ions, were almost absent. Let us now investigate this point. 

Ce+ ++ + 
Would changes in the ratio -vs-qppp affect the potential? If we made 

a large number of potential measurements, increasing in each experi- 
ment the concentration of the reduced form Ce +++ at the expense of the 
oxidized form Ce ++++ , we would find experimentally that the potential 
would fall in direct proportion to the decrease in this ratio. Evidently, 
then, the value of this ratio has an effect on the value of the potential 
developed. If now we wish to compare the potential developed by 
different oxidizing agents, we must choose some ratio as a standard; 
otherwise the comparisons will be valueless. The simplest ratio, 1, 
has been chosen. In our case, when the concentration of the eerie ion is 
equal to the concentration of the cerous ion, their ratio equals one. 
The potential developed by this cell is called the molar potential and is 
usually indicated by E . 



OXIDATION-REDUCTION REACTIONS 



147 



The outstanding usefulness of the molar potential can easily be 
shown. It enables us to catalogue all the oxidizing systems according 
to their oxidizing power. For instance, one can decide the following 
question: Which is the stronger oxidizing agent, eerie sulfate or potas- 
sium dichromate? We need merely look up their molar potentials in 

Ce + +"H- 
the appropriate tables. The value for -p +++ - = +1.45; that for 

-7T^+V = +1.3. This really means that under identical concentration 



conditions, the ability of the eerie ions to acquire electrons is greater 
than that of the dichromate ions, since its electrode becomes more posi- 
tive. Hence the eerie ion is a stronger oxidizing agent. 

TABLE XVIII. MOLAR POTENTIALS FOR OXIDIZING AGENTS 



Oxidized 
form 


Reduced 
form 


Eo 


Oxidized 
form 


Reduced 
form 


Ko 


MnOr 


Mn ++ 


+ 1.5 


Ou ++ 


Ou 


+0.34 


Ce ++++ 


Ce+++ 


+ 1.45 


Sn++++ 


Su-n- 


+0.14 


C1 2 


ci- 


+ 1.30 


HjO f 


II. 


0.0 


CrrfV 


(V f++ 


+ 1.3 


Sn +{ - 


Sn" 


-0.13 


Fe+H- 


Fc H 


+0.75 


Fo 1 -*- 


Fc 


-0.44 


I 2 


I- 


+0.54 


Zn f+ 


Zn 


-0.76 


Cu++ 


Cu+ 


+0.46 









READING REFERENCES 

GLASSTONE, S.: "Electrochemistry of Solutions," pp. S2&-359, 1). Van Nostrand 
Company, Inc., New York, 1930. 

HAMMKTT, L. P.: "Solutions of Electrolytes," 2d ed., ( -haps. VI and VII, McGraw- 
Hill Book Company, Inc., New York, 1936. 

LATIMER, \V. M.: "Oxidation Potentials," Prentice-Hall, Inc., New York, 1938. 

JETTE, E.: " Oxidation- Reduction Reactions," The Macmillan C'ornpany, New 
York, 1927. 

LATIMER, W. M., and J. H. HILDEBRAND: "Reference Book of Inorganic Chem- 
istry," rev. ed., pp. 471-481, The Macmillan Company, New York, 1940. 

QUESTIONS 

1. Balance the following equations by the ion-electron method: 
a. Hg 2 Cl 2 + aqua regia -> 
6. Mn++ + HN0 3 + NaBiOa -> 



148 QUALITATIVE ANALYSIS 



c. MnOr + H 2 2 - Mn++ + O 2 

d. SnS 8 - + Mg + H 8 O+ -* 
c. Cr 2 O 7 - + H 2 S + H 3 O+ -> 

/. VO++ + MnOr -> VOr + Mn++ 
g. S 2 8 - + C 2 0r -> SOr + C0 2 
h. lOr + 1- + H 3 O+ - I 2 + H 2 O 
i. SOr + H 8 O+ + Cu - 
j. S0 4 - + H 3 + + I" -> H 2 S + Ii + H 2 
k. P + OH- -> H 2 PO 2 - + PH 8 
i. Fe(CN) 6 H + H 2 O 2 -* FeCCNJe 181 4- H 2 O 
w. Mn ++ + Na 2 O 2 -> MnO 2 + Na + + OH- 

2. If you added Zn metal to a solution of Fe ++ , would oxidation-reduction take 
place? Write the equation. If metallic Fe were added to Zn ++ solution, what 
reaction would take place? 

3. Arrange the following in order of their oxidizing power: Ce +++ , Sn ++ , and 



4. Arrange the following in descending order of oxidizing power: Fe +++ , Cu ++ , 
Sn + +, and MnOr. 

6. How many grams of aluminum powder are needed to reduce 2 g. of KC1O 3 
toKCl? 

6. Write equations showing what would happen if H 2 S were passed into con- 
centrated HNOaj if it were passed in a solution of K 2 Cr 2 O?. 



PART II 

EXPERIMENTAL 



CHAPTER X 
INTRODUCTION 

One of the oldest branches of the science of chemistry is analysis. 
Its methods and techniques have been developed over a long period of 
time, from the crude furnaces and retorts of yesterday to the high-pre- 
cision instruments of today. Because of the limitations of the original 
techniquCvS, large quantities of material were required for analysis. 
With the improvement in methods, as time went on, the quantities 
were gradually decreased. Qualitative analysis followed this broad 
pattern and gradually became standardized so that it was customary to 
use from 15 to 25 ml. of solution for an analysis. 

With the advent of modern biochemical research and toxicology, 
the necessity for developing a new technique for the manipulation and 
analysis of the minute amounts of potent substances isolated from 
natural sources became apparent. As a result, a microtechnique was 
evolved, wherein 1 to 5 drops of a liquid or 2 to 6 mg. of a solid, could 
be handled successfully. This method was later extended to all the 
fields of chemistry. This technique is rather complex, however, both 
in the equipment and in the manipulative skill required of the analyst. 

The complexity of the technique led to a search for one that would 
combine the accuracy and speed of the micro methods with the simplic- 
ity of the classical methods and that could be taught successfully to the 
average undergraduate student. The solution was found in the semi- 
micro method. As used in qualitative analysis, the student uses from 
1 to 5 ml. of solution for an analysis. At the beginning of the course, 
the volume of solution taken for analysis is close to the higher figure; 
later, as the student's skill increases, the volume should approach the 
lower figure. Of course, allowance must be made for the complexity 
of the unknown sample. 

Semimicro qualitative analysis, as described in this book, uses the 
classical hydrogen sulfide scheme with some modifications. Some of 
the recently developed reactions using organic reagents are employed 
as confirmatory tests. This is not so radical an innovation as may 
appear, since organic reagents for nickel and cobalt have long been 
known and used. However, the indiscriminate use of organic reagents 
for most tests has been avoided, since experience has shown that many 

151 



152 QUALITATIVE ANALYSIS 

of the inorganic tests are equally sensitive and are better adapted for 
teaching the principles of inorganic chemistry. 

In keeping with the basic principles underlying semimicro analysis, 
the necessary apparatus has purposely been kept simple and the smaller 
sizes of standard equipment used as much as possible. The special 
items may now be purchased relatively inexpensively or may be con- 
structed by the student in his spare time. 

Although many varieties of apparatus found in the literature may 
be employed in semimicro qualitative analysis, the authors have found 
by experience that these generally exhibit no superiority to those 
described in the body of the book. 



CHAPTER XI 
MENTAL ATTITUDE 

Not so long ago, qualitative analysis, together with quantitative 
analysis, was definitely a " bread-and-butter " course, by the use of 
which most chemists earned their livelihood. Nowadays, the control of 
technical operations has to a large extent veered from the long-drawn- 
out chemical procedures to the shorter physical methods, such as the 
spectrographic and polarographic techniques. Qualitative analysis, 
far from having lost its importance as a teaching tool, now has acquired 
additional aims. Qualitative analysis has as its objectives the 
following: 

1. To furnish practical examples of the operations of the laws of 
chemistry. 

2. To furnish a body of facts pertaining to inorganic chemistry. 

3. To increase the manipulative skill of the student. 

4. To develop the scientific attitude. 

5. To serve as an introduction to the research ideology. 

A number of theoretical ideas have been presented to the student 
' regarding the fundamental laws and hypotheses of chemistry. The 
laboratory work in this text offers a practical demonstration of the 
application of many of these laws. In addition, the student has an 
opportunity to acquire a large number of facts concerning the chemistry 
of inorganic compounds. This entails the performance of a consider- 
able number of experiments, in the course of which the student has 
ample opportunity to increase and improve his manipulative skill and 
technique. 

The scientific attitude is based upon pertinent observations made 
in the course of experimental work, deductions drawn therefrom, and 
conclusions arrived at after coordination of all the evidence. Perti- 
nent observations are those pertaining to the formation, form, and 
color of precipitates; the evolution, color, and odor of gases; and any 
other changes apparent in the material under consideration. The art 
of observation may be acquired by the student. To further this aim, 
the preliminary experiments contain only a few references to the 
expected results. 

Yet, the mere gathering of observations is in itself but a beginning. 
The situation may be compared with that of a detective gathering 

153 



154 QUALITATIVE ANALYSIS 

clues. Experimental observations (like clues) are of no value unless 
the proper deduction can be drawn from them. For example, an 
observation by the student that his unknown solution is colorless 
should immediately lead to the deduction that colored ions, such as 
cupric, nickel, and chromate, are probably absent. If the solution is 
neutral, salts of the weak acid-strong base type, and vice versa, are also 
probably not present. 

The student should note carefully that the word probably is used. 
Any single piece of evidence is usually insufficient ground on which to 
base a conclusion. Let us refer once more to the example of the detec- 
tive; if he walks into a room, sees a man lying on the floor and a gun on 
the table, he has two clues (observations) and may therefore conclude 
that the man has been shot. This deduction may or may not be 
valid, however, and cannot be accepted as a conclusion until sup- 
porting evidence is obtained. Similarly, before drawing a conclusion 
in chemical analysis, the student must have a mass of evidence that is 
predominantly in favor of the conclusion he draws. Some of the 
deductions may point directly to the conclusion; others may be dia- 
metrically opposed; but on the whole, the evidence should be cumula- 
tive in the right direction. The conclusion must be reached on the 
basis of the impartial weighing of facts. The scientific attitude 
demands this impartial weighing of facts without permitting the intru- 
sion of preconceived notions and ideas. 

If the student adopts this attitude in investigating a troublesome 
point in the analysis, e.g., the unexpected appearance of a precipitate, 
he proceeds to make a series of experiments designed to disclose the 
nature of the precipitate. From the results obtained, the student 
may logically explain his difficulty. The student is now doing 
research. True, the point may be trivial and the facts well known, but 
from the student's point of view, it is research. Research, with a 
capital R, is merely the extension of this attitude and method to prob- 
lems lying at and beyond the boundary of present-day scientific knowl- 
edge. This point cannot be overstressed. Research is not a course 
that can be completed in so many hours but is the result of an attitude 
acquired by the student in the course of his chemical work. The 
research ideology extended to more complex problems, involves first, a 
search of the chemical literature to ascertain what is known and what 
has been done on the subject; then a series of experiments based on 
this knowledge is made, and certain deductions are drawn. More 
experiments are then carried out to test the validity of these deduc- 
tions. Finally, the evidence is weighed carefully and impartially and 
a conclusion drawn. 



MENTAL ATTITUDE 155 

Similarly, the student doing his unknowns should consider his 
laboratory manual and other more comprehensive textbooks as the 
recorded chemical knowledge for his purpose. On this basis, experi- 
ments are performed on the unknown and deductions drawn therefrom; 
the deductions are tested by further experimentation, and finally, 
conclusions are made as to the presence of certain substances. 

The authors have found that the analysis of solutions whose com- 
position is known, as a preliminary to analysis of the unknown, is 
unsatisfactory. This is so since, generally, the student does not then 
regard the "known" unknown as a -minor research problem but thinks 
of it rather as something to be done in cookbook fashion. This attitude 
is usually carried over to the analysis of the actual unknown. Also, 
in analyzing the known, the student may get a vague or indefinite test 
but does not check it, since he already knows whether the ion in ques- 
tion should be present. This may lead to an incorrect conclusion when 
the unknown is analyzed. In using the research idea, if the test is 
vague, the student will repeat the test, running simultaneously a con- 
trol test and a blank. 

A control test is one that is performed exactly as given in the analyt- 
ical procedure, except that a sample of the ion in question is used 
instead of the unknown. A blank test is a repetition of the given test, 
using all the reagents but omitting both the unknown and known 
samples of the ion. These two tests, together with the actual test on 
the unknown, should be run simultaneously and compared whenever 
the result is doubtful. 

The research ideology demands that the student be aware of the 
reason for each step in the analysis. Consequently, very detailed 
directions specifying exactly the amounts of this or that reagent or test 
solution are generally omitted. In many cases, however, the technique 
of the test involves the use of definite proportions of reactants, and in 
such cases detailed directions are, of course, given. 

The Notebook. One othor fundamental principle of the research 
ideology has not yet been mentioned: the keeping of an accurate and 
complete written record of the experimental work. Results should be 
recorded immediately in ink in a bound notebook. The student should 
never trust his results to his memory or to slips of paper. The note- 
book page should be divided into three vertical columns. In the first 
column should be placed, as briefly as is consistent with clarity, what 
has been done. In the second column is recorded the observation 
of the chemical change. Finally, in the last column, there is recorded 
an explanation that may consist of an equation, the formula of the 
product, a deduction, or a conclusion. An equation should be written 



156 



QUALITATIVE ANALYSIS 



for every unfamiliar or new reaction. It is not necessary for the student 
to keep an elaborate notebook. The notebook should be as brief as 
possible but should be so written that it will be understood by a person 
with equal chemical knowledge and enable him to draw the same con- 
clusions regarding an unknown. When an analysis has been com- 



Test- 



A/* 



Observation 



Explanation 



FIG. 42. Part of a notebook page. 

pleted, a report should be submitted listing briefly the conclusions 
arrived at. Of course, this implies that the completed notebook is 
available for examination at the same time. 

This chapter may be summarized with the statement that the crux 
of the course in qualitative analysis is the analysis of the unknown 
samples. These should be graduated in difficulty, and complexity and 
should serve as a test of the student's ability. 



CHAPTER XIII 
CATION ANALYSIS 

GROUP I CATIONS 

SILVER, Ag 

Silver occurs in Group 16 of the periodic table. Being compara- 
tively unreactive, it is found in the native state and has consequently 
been known from earliest times. It docs not displace hydrogen from 
acids or bases but is soluble in oxidizing agents, like nitric acid. 
Metallic silver is widely used in coins and common alloys. 

The silver cation, monovalent, forms more insoluble compounds 
than any other known metal. The nitrate and fluoride are soluble in 
water; the nitrite, acetate, sulfate, and chlorate are only moderately 
soluble. Chloride ion precipitates white silver chloride, AgCl, insolu- 
ble in nitric acid and hot water. It is appreciably soluble, however, in 
a large excess of chloride ion with the formation of the complex chloro- 
argentate (argentichloride) ions, [AgCl 3 ]~ and [AgCl 4 ] s . Silver chlo- 
ride is soluble in ammonium hydroxide, producing the complex 
diamminosilver ion, [Ag(NH 3 ) 2 ]+, from which AgCl may be reprecipi- 
tated by acidification. The chloride decomposes upon exposure to 
bright sunlight and in the presence of organic matter. The former 
characteristic property, given also by the bromide and iodide, is the 
basis of present-day photography. 

The alkali cyanides, KCN and NaCN, precipitate silver cyanide, 
AgCN, which is soluble in an excess of cyanide ion, forming the com- 
plex cyanoargentate (argenticyanide) ion, [Ag(CN) 2 ]~. The thiosul- 
fate ion, 8203", converts silver chloride to the complex [Ag 2 (S20 3 ) 3 ] s ; 
a reaction given by all insoluble silver salts. This explains the use of 
"hypo," Na 2 S 2 03, in the "fixing" of photographic films. Since metal- 
lic silver is rather low in the electromotive series, it may be displaced 
from solution by most of the common metals. The recovery of silver 
from thiosulfate wash solutions used in photography is accomplished in 
this manner. 

In neutral solution, ammonium hydroxide precipitates brown silvet 
oxide, Ag 2 0. Alkali carbonates yield white silver carbonate, AgaCOs, 
which slowly decomposes into Ag 2 and C02 on standing. Among the 

167 



168 QUALITATIVE ANALYSIS 

colored salts of silver are the pale yellow bromide, the yellow iodide 
and phosphate, and the orange f erricyanide. The black sulfide, Ag 2 S, 
is soluble in hot, dilute nitric acid. The red chromate, Ag 2 Cr04, and 
the red or red-violet precipitate obtained with an ammoniacal solution 
of p-dimethylamino benzalrhodanine are used as qualitative tests for 
the detection of silver. 

PRELIMINARY EXPERIMENTS 

1. To 1 ml. of Ag + test solution in a micro test tube, add dilute HC1 
dropwise until precipitation is complete. Centrifuge the tube and 
contents, and filter. Place a small amount of the precipitate on a spot 
plate, and expose to strong light for -a few minutes. 

Place another portion of the silver chloride on a slide with a black 
background, and add ammonium hydroxide carefully until solution is 
complete. Acidify the diammino silver complex with dilute nitric 
acid. Treat some of the AgCl with a slight excess of sodium thiosulfate 
solution. Ascertain the solubility of AgCl in hot water. 

2. Saturate 1 ml. of Ag + test solution with hydrogen sulfide. 1 
Filter and discard the clear filtrate. Test the solubility of the precipi- 
tate in cold, dilute HC1. To a second portion of the precipitate, add 
dilute HN0 3 , and heat carefully. Add more acid, if necessary, to effect 
complete solution. 

3. Using a capillary medicine dropper, place a few drops of Ag+ test 
solution on a piece of highly absorbent filter paper. From a second 
pipette, add an equal volume of 5 % potassium chromate solution to the 
test spot. 2 Repeat the experiment, using a micro test tube and the 
same proportions of reagents. 

4. To the slightly ammoniacal Ag + test solution on a spot plate, add 
a solution of p-dimethylamino benzalrhodanine (rhodanine reagent). 
Acidify with dilute nitric acid. 

5. (Optional) Detection of Ag in Coins or Valuable Alloys. Coins or 
other valuable alloys may be qualitatively tested for Ag without impair- 
ing the value or usefulness of the object. Rub the edge of a dime on a 
chip of unglazed porcelain. Moisten the line with a single drop of 
concentrated nitric acid, and warm carefully until the acid is completely 
evaporated. Carefully add a drop of the rhodanine reagent to the 
place where the line was traced. A violet-red colored zone is evidence 
of the presence of Ag in the coin. 

1 See Technique Filtration and Centrifugation, p. 158-159; Precipitation and 
H2S Generator, p. 157, 162. 

*See Technique Spot Tests, p. 164. 



GROUP I CATIONS 169 

LEAD, Pb 

Metallic lead falls in Group 4 of the periodic table. Although it is 
acted upon by all common acids and many weak organic acids, it is not 
readily dissolved, since many of the lead compounds thus formed arc 
insoluble. Since the nitrate is one of the most soluble salts, nitric acid 
is generally used to dissolve the metal. Lead finds use in the manufac- 
ture of low melting alloys. 

Lead forms divalent and tetravalent ions, the former being the 
more common and stable form. With chloride ion, white lead chloride, 
PbCl 2 , is obtained that is soluble in an excess of chloride ion with the 
formation of the complex anion, [PbCl 3 ]~. One gram of lead chloride is 
soluble in approximately 148 g. of H 2 O at 0C. and in approximately 
30 g. of H 2 at 100C. Therefore, lead can never be completely 
precipitated in Group I of the analytical scheme. Although the sulfate 
is very insoluble in water, it is soluble in ammonium acetate solution, 
because of the formation of slightly ionized lead acetate, Pb(C2H3O2)2. 
Lead ion gives a yellow precipitate with chromates, PbCr04, which is 
insoluble in acetic acid and water but soluble in nitric acid and alkalis. 

Tetravalent lead, in the form of lead dioxide, Pb02, is a good oxidiz- 
ing agent. Hence bcnzidine, which yields a blue compound when 
treated with oxidizing agents, may be used to detect load ion in the 
absence of other oxidizing agents. Hydrogen peroxide is used to 
convert divalent lead into lead dioxide. Alkali hydroxides precipitate 
white lead hydroxide, Pb(OH) 2 , from Pb++ test solutions. The 
hydroxide is soluble in excess reagent, forming the anion, [PbCh] 3 ". 

PRELIMINARY EXPERIMENTS 

1. Add dilute hydrochloric acid to 0.5 ml. of lead nitrate test solu- 
tion; filter and wash the precipitate with a few drops of cold water. 
(Test a portion of the wash water with H 2 S.) Treat a portion of the 
original precipitate with hot water, adding enough to ensure complete 
solution. Allow the solution to cool. Expose a small amount of the 
white lead chloride to strong light. Add a few drops of concentrated 
HC1 to another sample of the precipitate. 

2. Treat a few drops of the Pb ++ test solution with dilute H 2 SC>4. 
Test the solubility of the precipitate in hot water and in an excess of 
reagent. Add a few drops of glacial (concentrated) acetic acid and a 
few drops of a saturated solution of ammonium acetate to the lead 
sulfate, and warm gently. Add several drops of potassium chromate 
to the resulting solution. 



170 QUALITATIVE ANALYSIS 

3. Place a drop of Pb++ test solution on a filter paper, and add a 
mixture of equal volumes of dilute NH 4 OH and of 3% hydrogen perox- 
ide. Heat for a few minutes over a steam bath, and then add a drop of 
benzidine solution. 

MERCURY, Hg 

Mercury, the onlyirf^iaHic element that is liquid at ordinary tem- 
peratures, occurs in Groiiftx 2 of the periodic table. It forms two 
cations, the monovalent merdhrous, Hg 2 ++ , and the divalent mercuric, 
Hg+ + . The former is studied in this group and the latter in Group II 
of the analytical ^cheme. The metal itself is difficult to oxidize 
because of its low position in the electromotive series. It is easily 
soluble in hot nitric acid, giving the mercuric salt, and in cold nitric 
acid, forming the mercurous compound. The chloride, called calomel, 
is used medicinally as a purgative and externally as a mild antiseptic. 

The mercurous ion is stable to air but is converted to mercuric ion 
by strong oxidizing agents, like nitric acid. Mercurous ion gives a 
white precipitate with chloride ion, Hg 2 Cl 2 , which is insoluble in water 
and cold acids but is soluble, with oxidation to Hg ++ , in hot concen- 
trated HNO 3 , H 2 SO 4 , and aqua regia. With ammonium hydroxide, 
mercurous chloride undergoes an oxidation-reduction reaction, giving 
a mixture of black metallic mercury, Hg, and mercuric amido chloride, 
HgNH 2 Cl, which is white. With hydrogen s.ulfide, Hg 2 ++ forms black 
mercuric sulfide, HgS, and free mercury but no mercurous sulfide. 

Green mercurous iodide, Hg 2 I 2 , is obtained by the action of the 
iodide ion on mercurous salts; chromate ion, Cr0 4 == , gives a precipitate 
of red mercurous chromate, Hg 2 Cr() 4 , in boiling solution. Mercurous 
ion reacts with stannous chloride solution, SnCl 2 , to give a gray -black 
precipitate of elementary mercury and mercurous chloride, Hg 2 Cl 2 . 

PRELIMINARY EXPERIMENTS 

1. Add dilute HC1 to 1 ml. of Hg 2 ++ test solution until precipitation 
is complete. Ascertain the solubility of the precipitate in hot water. 
Test a portion of the precipitate for any change on exposure to light. 
To some of the precipitate, add cold dilute nitric acid. Treat another 
portion with aqua regia. 

2. Place the remainder of the precipitate on a spot plate, and care- 
fully add ammonium hydroxide. Ascertain the solubility of the 
precipitate in warm aqua regia. 

3. Pass H 2 S through a few drops of Hg 2 ++ test solution. Add 
potassium iodide solution to another sample of test solution. To a 
third portion, add potassium chromate solution, and heat to boiling. 



GROUP I CATIONS 171 

4. Suspend a few particles of mercurous chloride in water, and add a 
slight excess of stannous chloride solution. 

ANALYSIS OF GROUP I CATIONS 
OUTLINE OF SEPARATIONS 

The three cations of this group are precipitated as the sparingly 
soluble or insoluble chlorides. Lead chloride is separated from the 
group precipitate because of its complete solubility in hot water. 
Silver and mercurous chlorides are separated by their reaction with 
ammonium hydroxide, the former yielding a soluble complex ion, the 
latter a mixture of elementary mercury and mercuric amido chloride. 

ANALYTICAL PROCEDURE 

1. Place 2 ml. of the neutral or slightly acid unknown solution in a 
micro test tube, and add HC1 until precipitation is complete. Filter 
and wash the precipitate once with the least amount of cold water 
necessary. Reserve the combined filtrate and washings for the analysis 
of succeeding groups. 

NOTE: If the solution is strongly acid, add NH 4 OIl until slight precipi- 
tate forms, then add a few drops of HNO 3 to clear the solution, before 
adding HG1. 

2. Wash the precipitate once with hot water, and filter. Test for 
lead by placing a drop of the filtrate on a spot plate and adding an 
equal volume of potassium chromate solution. 

A yellow precipitate, insoluble in dilute acetic acid, indicates 

lead. 

Place another drop of the filtrate on a piece of filter paper, and add a 
few drops of a previously mixed solution of equal volumes of dilute 
NH 4 OH and 3% H 2 2 . Heat the paper over a steam bath for a 
minute or two, and add a drop of benzidine solution to the wet spot. 

A blue color proves the presence of lead. 

If lead has been found, extract the precipitate exhaustively with hot 
water. 

3. To the residue, from step 2, add ammonium hydroxide and stir. 
Filter and acidify a portion of the filtrate with dilute nitric acid. 

A white precipitate indicates silver. 



172 



QUALITATIVE ANALYSIS 



Place a drop of the filtrate on a spot plate, and add an equal volume of 
p-dimethylamino benzalrhodanine (rhodanine) solution. Acidify with 
HNO 3 . 

A red or red- violet precipitate confirms the presence of silver. 

NOTE: In the presence of large amounts of Hg 2 Cl 2 and very little AgCl, 
the latter may not dissolve in NHjOH. The mercurous compound may 
be oxidized to the soluble mercuric salt by treating the black residue, after 
the NH 4 OH treatment, with a mixture of 1 part of HC1 to 3 parts of 
bromine water. The residue is finally washed with hot water and then 
treated with NH^OH. The solution is then tested for Agt*~ as above. 

If silver has been found, treat the precipitate thoroughly with NH 4 OH 
until no further test for Ag+ is obtained. 

A black residue indicates the presence of mercury. 

4. Add a few drops of aqua regia to the black precipitate, and warm 
until it is all dissolved. Add an excess of SnCl 2 solution. 

A white precipitate turning gray confirms mercury. 

ANALYTICAL TABLE I 
ANALYSIS OF GROUP I 

Unknown solution may contain the cations of all five analytical groups. Treat 
with HC1: white precipitate of PbCh, Hg 2 Cl 2 , and AgCl. Filter, reserve filtrate 
for subsequent group analyses (1).* Extract precipitate with hot H 2 (2). 



FUtrate : Pb++. Test por- 
tion with CrC^"" solu- 
tion. Confirm with 
benzidine test (2). 


Precipitate: AgCl, Hg 2 Cl,. Treat with NH 4 OH, stir 
and filter (3). 


Filtrate: Ag(NH 8 ) 2 + . 
Acidify portion with di- 
lute HNO 3 . Confirm 
Ag + with Rhodanine test 
(3). 


Precipitate : Hg (black) and 
HgNH 2 Cl (white). Dis- 
solve in aqua regia and 
test with SnCl 2 solution 
(4). 



* Boldfaced numbers refer to steps in the procedure. 

GROUP II CATIONS 

MERCURY, Hg 

The divalent mercuric ioh, Hg' 1 " 4 ", forms a soluble chloride. With 
ammonium hydroxide, HgCU yields a white precipitate of mercuric 
amido chloride, HgNH 2 Cl. Hydrogen sulfide precipitates black 
mercuric sulfide, HgS, which is insoluble in water, dilute acids, con- 
centrated hydrochloric acid, and ammonium polysulfide, (NH^S*, but 
is soluble in aqua regia and, after long boiling, in concentrated nitric 
acid. Metals above mercury in the electromotive series displace the 
metal from its salts. 



GROUP II CATIONS 173 

Alkali hydroxides precipitate yellow or red mercuric oxide, HgO, 
which is insoluble in an excess of reagent. Chromate ion precipitates 
orange mercuric chromate, HgCrO 4 ; the iodide ion precipitates red 
mercuric iodide, Hgl2, which is soluble in excess reagent with the 
formation of the colorless complex ion, [HgL] 553 . Stannous chloride 
reacts with mercuric chloride to form first a white precipitate of 
Hg2Ch, which changes to free mercury upon the addition of excess 
reagent. Diphenyl carbazide, added to an acid solution of Hg ++ gives 
a violet or blue precipitate. 

Mercuric sulfide, in ointment form, is used in the treatment of cer- 
tain skin conditions. Amalgams are used in dentistry. Even slight 
concentrations of mercuric salts, taken internally, aro violent poisons. 

PRELIMINARY EXPERIMENTS 

1. Acidify 1 ml. of Hg ++ test solution with dilute HC1, and saturate 
with H 2 S. Filter and wash the precipitate with warm water. Treat a 
portion of the black precipitate with aqua regia. Treat another por- 
tion with dilute HN0 3 , and heat just to boiling. Repeat the experi- 
ment, boiling the mixture for several minutes. Ascertain the solubility 
of the precipitate in (NH^S*. 

2. Add ammonium hydroxide carefully to a few drops of HgCl 2 
solution. 

3. To the Hg ++ test solution add very slowly a solution of stannous 
chloride until an excess has been added. Place a drop of Kg** test 
solution on a filter paper that has been moistened with a solution of 
diphenyl carbazide in ethyl alcohol. 

4. (Optional) Add either KOH or NaOH to a few drops of Hg+ + 
test solution ; then treat with an excess of reagent. To another portion 
of the test solution, add some KI solution. Ascertain the effect of 
excess reagent on the precipitate. Treat a third portion of Kg 4 "*" test 
solution with K 2 CrO4 solution. 

LEAD, Pb 

The sulfide ion precipitates black lead sulfide, PbS, from slightly 
acid, neutral, or weakly alkaline solutions. The sulfide is insoluble in 
water, dilute acids, bases, carbonates, and alkali sulfides. It is soluble 
in dilute nitric acid with the separation of elementary sulfur and in 
concentrated hydrochloric acid (see Lead, Group I). 

PRELIMINARY EXPERIMENTS 

1. Acidify 1 ml. of Pb++ test solution with dilute HC1, warm 
slightly, and saturate with H 2 S. Filter and wash the precipitate, and 



174 , QUALITATIVE ANALYSIS 

treat a portion of it with dilute HN0 3 . Add several drops of dilute 
H2S04 to the solution, and evaporate to dense fumes of S0 3 . After 
cooling, pour the solution carefully into water. Filter the precipi- 
tate, and treat it with a warm, saturated solution of ammonium acetate 
that has been acidified with acetic acid. To the resulting solution, add 
K 2 Cr0 4 solution. 

2. Repeat the benzidine test for lead as given under Group I. 

BISMUTH, Bi 

Bismuth falls in Group 5 of the periodic table. The metal is 
rather inactive chemically, being below hydrogen but above mercury 
in the electromotive series. It is insoluble in hydrochloric acid and 
cold sulfuric acid but is soluble in nitric acid and hot sulfuric acid. 

The cation usually exhibits a valence of 3. Most of the bismuth 
salts are insoluble. The soluble salts hydrolyze in water with the for- 
mation of the hydroxy compounds. Thus bismuth trichloride, BiCl 3 , is 
converted to bismuth dihydroxychloride, Bi(OH) 2 Cl, by water. This 
hydrolysis is inhibited by the presence of the corresponding free acid. 
The 'dihydroxychloride is therefore soluble in concentrated hydro- 
chloric acid, reforming the trichloride. On long standing, the hydroxy- 
chloride is converted to the difficultly soluble oxyehloride, BiOCl. 
The trichloride and dihydroxychloride form dark brown, insoluble 
bismuth sulfide, BioSs, with sulfide ion. This precipitate is insoluble in 
water and ammonium polysulfide but dissolves in hot hydrochloric 
acid and in dilute nitric acid. 

Alkali hydroxides precipitate white bismuth hydroxide, Bi(OH) 3 , 
which is insoluble in water and excess reagent but is soluble in acids. 
Alkali stannites, like sodium stannite, Na 2 Sn0 2 , added to bismuth 
compounds, precipitate black, metallic bismuth. Cinchonine reagent, 
in faintly acid solution, gives an orange precipitate with bismuth ion. 

PRELIMINARY EXPERIMENTS 

1. Saturate 1 ml. of an acid solution of Bi +++ with H 2 S. Filter and 
wash the precipitate with water. Ascertain the solubility of the sulfide 
in ammonium polysulfide. Treat some of the precipitate with dilute 
HN0 3 , and boil for a short time. Add a few drops of concentrated 
H2SO4 to the solution, and evaporate to fumes of sulfur trioxide. After 
cooling, pour the solution into cold water, and add excess NHUOH^ 
Dissolve a small portion of the precipitate in concentrated HC1. 
Treat several drops of the solution with cold water. Test the solubility 
of the precipitate in an excess of HC1. 



GROUP II CATIONS 175 

2. Dissolve some Bi(OH)3 in the least amount of HC1 necessary. 
Add 1 drop of the solution to filter paper previously impregnated with 
1 drop of concentrated NH 4 OH. Add 1 drop of freshly prepared 
sodium stannite solution to the mixture on the paper. 

3. Place a drop of the faintly acid Bi+ ++ test solution on filter paper 
that has been previously treated with a drop of cinchonine reagent. 

COPPER, Cu 

Copper is placed in Group 1 of the periodic table. The element 
resembles mercury in some of its reactions. It is soluble in oxidizing 
acids like nitric acid and hot, concentrated sulfuric acid. Since it is 
above both mercury and silver in the electromotive series, it may dis- 
place both these cations from their compounds. Copper is an excellent 
conductor of heat and electricity. Because copper salts are poisonous, 
they are often used commercially as fungicides. 

Copper forms two cations, the monovalent cuprous and the divalent 
cupric ions. The latter are more common analytically and arc there- 
fore the only type considered. Cupric salts are either blue or green in 
dilute aqueous solution; in the anhydrous state, they are usually 
either white or yellow. Ammonium hydroxide, in small amounts, 
precipitates basic salts, like Cu2(OH) 2 S()4, which are extremely soluble 
in an excess of reagent, forming a deep blue solution of the tetrammino- 
cupric ion, [Cu(NH 3 )4] ++ . Alkali cyanides convert this ion to the 
colorless cuprocyanide ion, Cu(CN) 3 =s . This ion is only very slightly 
dissociated and does not give a precipitate of copper sulfide on treat- 
ment with hydrogen sulfide. 

From solutions of Cu ++ , hydrogen sulfide precipitates black cupric 
sulfide, CuS, which is insoluble in dilute, nonoxidizing acids and 
slightly soluble in ammonium polysulfide, (NH^S^. The sulfide is 
soluble in hot, dilute nitric acid. It is also soluble in alkali cyanides, 
the complex cuprocyanide ion being formed with the latter reagent. 
Sodium and potassium hydroxides precipitate blue cupric hydroxide, 
Cu(OH) 2 , from Cu 4 " 1 " solutions. The hydroxide is converted to brown- 
black cupric oxide, CuO, on standing or boiling. 

Potassium ferrocyanide, K 4 Fe(CN) 6 , precipitates red-brown cupric 
ferrocyanide, Cu 2 Fe(CN) 6 , insoluble in dilute acids but soluble in 
ammonium hydroxide. A green precipitate is formed in the Cu ++ spot 
test with a-benzoinoxime. 

PRELIMINARY EXPERIMENTS 

1. Saturate 1 ml. of a warm, acidified solution of Cu++ with H 2 S. 
Filter and wash the precipitate with hot water. Boil a portion of the 



176 QUALITATIVE ANALYSIS 

precipitate with dilute HNOs. Cool the solution, add NaOH, and 
heat. 

2. Treat another portion of the test solution carefully with NH 4 OH ; 
then add an excess of reagent. To a few drops of the ammoniacal solu- 
tion, add KCN drop wise until the blue color just disappears, then add 
2 drops in excess. (CAUTION: Never acidify cyanide solutions.) 
Saturate the colorless solution with H 2 S. 

3. Acidify a portion of the test solution with dilute acetic acid, and 
then add K 4 Fe(CN) 6 solution. Test the solubility of the precipitate 
in both NH 4 OH and dilute HN0 3 . 

4. Place a drop of the Cu ++ test solution on a piece of filter paper, 
and add an equal volume of a 5 % solution of a-benzoinoxime to the wet 
spot. Hold the paper over an open bottle of NH 4 OH for several 
minutes. 

5. (Optional) Immerse a penny in a small volume of concentrated 
HNOa for about 5 sec. ^Cautiously make the solution strongly 
ammoniacal. 

CADMIUM, Cd 

Cadmium is found in Group 2 of the periodic table. It is moder- 
ately active, being a little above hydrogen in the electromotive series. 
It is readily soluble in nitric acid and dissolves slowly in hydrochloric 
acid and sulfuric acid. Cadmium burns in air to form the brown oxide, 
CdO. The metal is used extensively in the manufacture of yellow 
paint and also finds great use in the manufacture of a number of valua- 
ble low-melting alloys. 

The cation is divalent in all its compounds. Sulfide ion precipitates 
yellow or orange cadmium sulfide, CdS, which is insoluble in water, 
ammonium hydroxide, ammonium polysulfide, and alkali cyanides. 
The sulfide is soluble in hydrochloric acid, nitric acid, and hot, dilute 
sulfuric acid. Alkali hydroxides yield white cadmium hydroxide, 
Cd(OH) 2 , which is insoluble in excess reagent but is soluble in acids. 
Ammonium hydroxide also precipitates Cd(OH) 2 , which is soluble in 
an excess of reagent with the formation of the colorless [Cd(NH 3 ) 4 ] + + 
ion. This ion is converted to the colorless cadmocyanide (tetra- 
cyanocadmate) ion, [Cd(CN) 4 ]" =s , upon reaction with alkali cyanides. 
However, this ion dissociates to some extent, furnishing sufficient 
cadmium ions to precipitate cadmium sulfide with hydrogen sulfide. 

Potassium ferrocyanide forms white cadmium ferrocyanide, 
Cd 2 Fe(CN) 6 , with Cd+ + test solutions. A yellow precipitate 
or solution is obtained upon the reaction of cadmium solutions with 
thiosinamine . 



GROUP II CATIONS 177 

PRELIMINARY EXPERIMENTS 

1. Saturate 1 ml. of Cd ++ test solution, acidified with dilute acetic 
acid, with H^S. Hoat to boiling, filter, and wash the precipitate. 
Ascertain the solubility of the precipitate in hot, dilute HNOs. Make 
the resulting solution alkaline with NaOH. 

2. Repeat the preceding experiment, strongly acidifying the test 
solution with HC1. 

3. Add NH 4 OH carefully to another portion of test solution, using 
enough excess to redissolve the precipitate that first forms. To the 
solution, add KCN solution, and then saturate with H 2 S. (CAUTION.) 

4. Add some K 4 Fe(CN) 6 solution to a few drops of the test solution. 

5. Make a few drops of Cd ++ test solution alkaline with NaOH. 
Add a small crystal of thiosinamine, and warm. 

ARSENIC, As 

Arsenic, which falls in Group 5 of the periodic table, exists in several 
allotropic modifications; the most usual form is steel-gray in color and 
has a metallic luster. Since it is in Group 5, it is very similar to 
phosphorus and forms corresponding series of salts; arsenites, 
AsOif, and arsenates, As04~. It is, however, more metallic than 
phosphorus and therefore forms simple cations, trivalent arsenous 
and pentavalent arsenic. 

Arsenic dissolves readily in nitric acid. It is insoluble in cold, 
concentrated hydrochloric acid and cold, dilute sulfuric acid. The 
metal, on heating in air, forms arsenic trioxido, As 2 O 3 , with the simul- 
taneous production of a garlic odor. The vapors are poisonous. 
Arsenic compounds have been known to man since ancient times. 
In modern times, they are used extensively as poisons in fungicides, 
wood preservatives, and poison gases like Lewisite. Organic arsenic 
compounds are also used medicinally as spirocheticidal and trypano- 
cidal agents. 

The trioxide is only very slightly soluble in water but dissolves 
readily in hydrochloric acid with the formation of the arsenous ion, 
As +++ . With alkali hydroxides, the oxide forms the arsenite ion, 
As0 2 ~~. The pentoxide, As 2 5 , is easily soluble in water, forming 
arsenic acid, H 3 AsO 4 ; with alkalis it produces arsenates. 

Hydrogen sulfide readily precipitates the yellow trisiilfide from 
0.3JV acid solutions of arsenous ion. Under the same conditions, how- 
ever, arsenic pentasulfide is not precipitated from arsenate solutions, 
only the trisulfide being obtained after long treatment with hydrogen 
sulfide. The pentasulfide may be precipitated from a cold solution 



178, QUALITATIVE ANALYSIS 

containing a large excess of concentrated hydrochloric acid. A mixture 
of the two sulfides is precipitated when a hot solution of arsenate ion in 
concentrated hydrochloric acid is treated with hydrogen sulfide. In 
neutral or alkaline solution, soluble thio salts are obtained on treatment 
with hydrogen sulfide. Similar salts are formed on treatment of the 
sulfides with alkali sulfides and poly sulfides. Arsenic trisulfide, with 
ammonium sulfide, (NH 4 ) 2 S, gives the thioarsenite ion, AsS 3 s . With 
ammonium polysulfide, (NH 4 ) 2 S X , oxidation and solution takes place 
with the formation of the thioarsenate ion, AsS4 s . Arsenic penta- 
sulfide yields only the thioarsenate ion under the same conditions. 
Neutralization of the solutions produces the trisulfide from the thio- 
arsenite and the pentasulfide from the thioarsenate. Arsenic penta- 
sulfide is soluble in ammoniacal H 2 02 with the formation of As0 4 =. 

Neutral solutions of arsenite ion yield, with silver nitrate, yellow 
silver arsenite, AgsAsOa, which is soluble in dilute acids and ammonium 
hydroxide. Arsenates under the same conditions form chocolate 
brown silver .arsenate, Ag 3 As0 4 , soluble in the same reagents. Treat- 
ment of arsenites with stannous chloride in concentrated hydrochloric 
acid solution produces a brown to black precipitate of metallic arsenic. 
This is the Bettendorf test. 

Reduction of arsenic compounds with acids and active metals 
produces arsine, AsH 3 . This gas is used to identify arsenic in several 
ways. One of the most important, the Gutzeit test, depends upon the 
reaction of arsine with a crystal of silver nitrate; a yellow coloration of 
AsAg 3 -3AgN0 3 is obtained that rapidly turns to black, metallic silver. 
Treatment of trivalent arsenic compounds with hydrochloric acid and a 
strip of bright copper foil leads to the deposition of a gray film of CusAs2 
on the surface of the metal. This is the Reinsch test. 

Arsenates give a white precipitate of magnesium ammonium 
arsenate, MgNH 4 As0 4 , with magnesia mixture. Arsenites do not 
react. With ammonium molybdate, (NH 4 )2MoO 4 , arsenates form a 
yellow precipitate of ammonium arsenomolybdate, (NH 4 ) 3 AsO 4 - 
12MoOa. Sodium thiosulfate precipitates arsenic pentasulfide from 
faintly acid solution. 

PRELIMINARY EXPERIMENTS 

1. Make 1 ml. of AS+++ test solution Q.3N with HCL Saturate the 
solution with H 2 S at room temperature. 

NOTE: To make a solution 0.3#, neutralize the solution, and then add 
the amount of acid needed to bring the solution to 0.3#. 

2. Repeat the experiment, using pentavalent arsenic test solution. 
Repeat, using a hot solution. 



GROUP II CATIONS 



179 



. Crystal of 

silver nitratt 



-Filter 
peper 



3. Make 1 ml. of As 4 " 5 test solution 0.5AT with HC1, heat the solution 
to boiling, and saturate with H 2 S. Filter and wash the precipitate. 
Add (NHOuSs solution to one portion of the precipitate. Save the 
rest for step 4. Neutralize the solution with dilute HC1 and filter. 
Add concentrated HC1 to the precipitate and heat to boiling. Ascer- 
tain the solubility of the precipitate in NH 4 OH and NaOH. 

4. Place a small amount of As 2 S 5 precipitate in a crucible, and dis- 
solve it in the least volume of ammoniacal H 2 2 required. Add 
magnesia mixture, evaporate to dryness, and ignite. Cool the crucible, 
and moisten the residue with a few drops of Bettendorf 's reagent. 

5. Treat a few drops of arsenate test solution with AgNO 3 solution. 
Repeat, using a neutral arsenite test solution. Add magnesia mixture 
to several drops of both arsenate and 

arsenite test solutions. Treat separate 
portions of the test solutions with a 
hot HN0 3 solution of ammonium 
molybdate. 

6. Gutzeit Test. In a test tube, 
place a small amount of arsenic test 
solution; add a stick of As-free Zn and 
a small volume of dilute H 2 S0 4 . Place 
a wad of cotton in the mouth of the 
tube, and fold a piece of filter paper 
with a crystal of AgNOaOver the mouth 
of the tube. Run the test in conjunc- 
tion with a blank, using Zn and acid 
alone. 

7. Place a piece of copper foil in a 
test tube, add a few drops of an HC1 
solution of As***, and warm. 

8. (Optional) On a silver coin previously cleaned with HN0 3 , place 
a drop of As+ ++ test solution acidified with HC1. Touch the coin and 
liquid with a stick of As-free zinc. After the black stain has developed, 
remove the zinc, and wash the coin with water. Ascertain the solubil- 
ity of the stain in NaOBr. 

ANTIMONY, Sb 

Antimony, which falls in Group 5 of the periodic table, is a rather 
unreactive element. The metal has a bright silvery luster and does not 
tarnish readily in dry air. It has been known from early times. At 
present, it is used chiefly in alloys, storage-battery plates, and 
medicinals. 




Fro. 52. Gutzeit test. 



180 QUALITATIVE ANALYSIS 

Since it is in the same family as arsenic, it forms cations exhibiting 
valences of 3 and 5, and the anions Sb0 3 s and Sb(>4-. However, hav- 
ing a higher atomic number than arsenic, it is more metallic in char- 
acter. The trivalent ion is slightly basic in its reactions. 

Trivalent antimony salts, like the chloride, are readily hydrolyzed 
in water, precipitating white hydroxy salts like the dihydroxy chloride, 
Sb(OH) 2 Cl. This precipitate, when fresh, is readily soluble in con- 
centrated mineral acids, tartaric, citric, and oxalic acids but not in 
acetic acid. Alkalis and alkali carbonates precipitate white Sb(OH)s, 
which is soluble in acids or an excess of strong alkali. 

From trivalent antimony solutions or from freshly precipitated 
Sb(OH)2Cl, sulfide ion precipitates orange-red antimony trisulfide, 
Sb2S 3 . The sulfide is insoluble in dilute, nonoxidizing acids but is 
soluble in concentrated hydrochloric acid. It is also soluble in 
ammonium polysulfide with oxidation and the formation of a soluble 
thio derivative of antimonic acid, SbS^. 

Pentavalent antimony ion forms, with sulfide ion, orange antimony 
pentasulfide, Sb 2 S 6 , which is soluble in ammonium polysulfide with the 
formation of the thioantimonate ion. When the solution is treated 
with dilute hydrochloric acid, Sb 2 S 4 is precipitated; this is soluble in 
concentrated hydrochloric acid. 

Antimony is below hydrogen in the electromotive series and is 
therefore displaced from its salts by many metals. This fact is utilized 
in the "couple test," in which bimetallic couples like silver and tin, 
copper and tin, zinc and silver, and platinum and zinc are used. 
Antimony ion, in contact with such a couple, is plated out on the nobler 
metal. The metal, unlike arsenic, is only slowly soluble in sodium 
hypobromite solution. With sodium thiosulfate in hot, slightly acid 
solution, antimony ion precipitates as "antimony cinnabar," SbOS2. 
Rhodamine B gives a violet color with pentavalent antimony solutions. 

PRELIMINARY EXPERIMENTS 

1. Make 1 ml. of Sb +++ test solution 0.3N with HC1. Saturate with 
EUS ; filter and wash the precipitate with hot water. Treat a portion of 
the precipitate with (NH^S*, using enough to effect complete solution. 
Acidify the resulting solution with dilute HC1. Treat the precipitate 
with concentrated HC1. Warm to effect solution and to expel excess 
H2S. Dilute the solution with an equal volume of water, and pass in 
H 2 S. 

2. Place a few drops of the acid test solution on a silver coin. 
Touch the liquid and metal with a piece of tin. Wash the spot with 
water, and add a drop of NaOBr solution. 



GROUP- II CATIONS 181 

3. Add H20 to a few drops of Sb++ + test solution until a precipitate 
forms. Treat the suspended solid with concentrated HC1. 

4. Antimony Cinnabar Test. Add dilute NH 4 OH solution to the 
test solution until precipitation just begins. Heat to boiling, and add 
a few crystals of Na 2 S 2 3 . 

5. To a portion of the test solution add a small amount of solid 
NaN0 2 or KN0 2 . Shake until the evolution of gas ceases. Fill one 
of the depressions in a spot plate almost to the top with dilute rhoda- 
mine B solution. To this add 2 drops of the oxidized antimony 
solution. 

TIN, Sn 

Tin is a soft, malleable metal, known since prehistoric times. It is 
placed in Group 4 of the periodic table. Being just above hydrogen in 
the electromotive series, it displaces the latter from most acids. 
However, with nitric acid, it yields white, insoluble metastannic acid. 
Tin is used chiefly in the manufacture of alloys solder, brass, bronze, 
type metal, Wood's metal and in plating. 

Tin forms divalent (stannous) and tetravalent (stannic) cations and 
the corresponding anions, stannite, Sn0 2 , and stannate, SnOs"*. The 
stannous ion is more basic in its reactions than the stannic form. The 
latter form is also more stable, stannous ion undergoing oxidation in air 
to the higher valence state. 

Alkalis and alkali carbonates precipitate white stannous hydroxide, 
Sn(OH) 2 , which is soluble in excess alkali with the formation of the 
metastannite ion, (HSnO a )~~. On standing, solutions of stannite ion 
slowly decompose into brown stannous oxide, SnO, and metallic tin. 
Stannous salts hydrolyze in water, starmouH chloride forming first 
white hydroxychloride, Sn(OH)Cl, and then the hydroxide, with an 
excess of water. The addition of hydrochloric acid and metallic tin to 
stannous ion solutions prevents hydrolysis of the aqueous solution and 
oxidation to the stannic ion. 

In not too acid solution, sulfide ion precipitates brown stannous 
sulfide, SnS, from stannous ion solutions. The sulfido is soluble in 
concentrated hydrochloric acid but is insoluble in ammonium hydrox- 
ide. With ammonium polysulfide, stannous sulfide dissolves, with 
oxidation, to form the thiostannate ion, SnSs"". Acids precipitate 
yellow stannic sulfide, SnS 2 , from this solution. From solutions of 
stannic ion, sulfide ion precipitates yellow stannic sulfide slowly. It is 
soluble in hydrochloric acid, ammonium polysulfide, and sodium and 
potassium hydroxides. 



182 QUALITATIVE ANALYSIS 

Stannous and stannite ions are excellent reducing agents. With 
mercuric chloride, stannous chloride forms white mercurous chloride 
and then black, metallic mercury, with an excess of reagent. Stannic 
chloride does not react with mercuric chloride. Alkali stannites, 
freshly prepared, reduce bismuth ion to metallic bismuth. Stannous 
ion produces a violet color with cacotheline. Stannic compounds must 
first be reduced with magnesium and acid before applying the test. 

4 

PRELIMINARY EXPERIMENTS 

1. Pass H2S into a faintly acid test solution of Sn** until precipita- 
tion is complete. Filter and wash the precipitate. Treat some of the 
precipitate with orange (NH^S* solution, warming to 60C., if neces- 
sary. Acidify the solution with dilute HC1. Add concentrated HC1 
to another portion of the precipitate, and heat. Saturate the solution 
with H 2 S. / 

2. Repeat all the foregoing, using Sn ++++ test solution. 

3. To several drops of SnCl2, in a test tube, add a fow drops of 
HgCl 2 , followed by an excess of reagent. Repeat the test with SriCl 4 . 

4. Place a 2-cm. strip of magnesium ribbon into 1 ml. of an acid 
solution of SnCU. Add more acid, if necessary, to dissolve the metal 
completely. Filter rapidly, and test the filtrate with HgCl 2 . 

5. Place 1 or 2 drops of an acid solution of Sn ++ on a piece of filter 
paper. Treat the wet spot with an equal volume of cacotheline 
solution. 

ANALYSIS OF GROUP II CATIONS 

OUTLINE OF SEPARATIONS 

The cations of Group II precipitate as sulfides in acid concentrations 
of 0.3N with respect to hydrochloric acid, when treated with hydrogen 
sulfide. At this pH, the sulfide ion concentration is sufficiently 
repressed to prevent precipitation of the Group III cations. The 
cations are further divided into two subgroups, A and 5, on the basis of 
their solubility in ammonium polysulfide. Subgroup A, the insoluble 
sulfides, are mercuric mercury, lead, bismuth, copper, and cadmium; 
subgroup B, the soluble sulfides, are arsenic, antimony, and tin. 

The sulfides of subgroup A are separated as follows: mercuric sulfide 
by its insolubility in nitric acid; lead by the insolubility of its sulfate; 
bismuth as the insoluble hydroxide. Copper and cadmium are con- 
verted to the cyanide complexes and separated with hydrogen sulfide, 
the former giving no precipitate because of the stability of the cupro- 
cyanide ion. 



GROUP II CATIONS 183 

The sulfides of subgroup B are reprecipitated with hydrochloric 
acid. Arsenic is separated because of the insolubility of its sulfide in 
concentrated hydrochloric acid. Antimony and tin are detected in the 
presence of each other. 

ANALYTICAL PROCEDURE 

The unknown may be the filtrate from Group I or an unknown solu- 
tion containing the mixed cations of Group II and subsequent groups. 

1. In either case, adjust the acidity to 0.3AT, and precipitate the 
sulfides as follows. Neutralize the unknown with NH 4 OH, and then 
make faintly acid by adding 1 drop of very dilute HC1. To 3 ml. of 
this solution add 5 drops of 6N HC1, heat the solution to boiling and 
saturate with HoS. Cool, dilute to 5 ml. with distilled water, and pass 
in H 2 S again until precipitation is complete. 

2. Filter and wash the precipitate with hot water. Save the pre- 
cipitate for step 3. Combine the filtrate and washings, make strongly 
acid with HC1, and boil till all the H 2 S is expellod. Discard the residue, 
and save the filtrate for the analysis of Groups III, IV, and V. 

NOTE: Tt in often advisable to wash the precipitate with 0.3Af HC1, 
which has been saturated with H 2 S. 

3. Extract the precipitate from step 2 once or twice with small 
portions of fresh ammonium polysulfide heated to about 60C. 1 
Filter and reserve the filtrate for the analysis of subgroup B (step 11). 

SUHGUOUP A 

4. Heat the precipitate remaining from the preceding treatment 
almost to boiling with about 0.5 ml. of dilute nitric acid. Filter and 
wash the precipitate with cold water. Save the filtrate for step 6. 

5. Dissolve a portion of the black precipitate of HgS from step 4 in 
aqua regia. Evaporate nearly to dryness (moist solid) to expel the 
excess reagent. Dilute with several drops of water, and add a slight 
excess of SnCU solution. 

White precipitate turning gray-black indicates mercury. 

To a second portion of the precipitate, add about 0.5 ml. of bromine 
water, and heat on a water bath until the solution becomes colorless. 
Repeat with another 0.5 ml. of bromine water. Add 5 drops of con- 
centrated H 2 SO 4 to this solution, and evaporate to copious fumes of 
S0 3 (use a crucible). Cool, dilute the solution^ with several drops of 

1 See Technique, Extraction, p. 161. 



184 QUALITATIVE ANALYSIS 

water, and add an excess of solid sodium acetate. Treat with several 
drops of diphenyl carbazide solution and 1 ml. of amyl alcohol. 

Deep blue coloration in the amyl alcohol layer proves presence 

of mercury. 

NOTE: The success of this test depends upon the complete conversion of 
the mercuric sulfide to the bromide and upon the subsequent elimination 
of all of the bromide ion as HBr. Mercuric halides, because of their slight 
ionization, give only very weak tests with diphenyl carbazide. It is 
advisable always to run a parallel control experiment with known HgS. 

6. Boil the filtrate from step 4 with about 0.5 ml. of concentrated 
[2804 until dense white fumes of SOs are given off. Cool the solution, 
and pour cautiously into 1 to 2 ml. of cold water. Filter and save 
the filtrate for step 7. 

NOTE: If considerable amounts of bismuth are present, some (BiO)2S2O 7 * 
SH^O may also be precipitated which may be recognized by its coarser 
crystals. To correct this condition, treat the precipitate with 1 ml. of 
dilute HC1 and 0.5 ml. of concentrated HaSCh, heat to copious fumes of 

SO 3 , and proceed as before. 

* 

Heat the precipitate of PbS0 4 almost to boiling with about 0.5 ml. of 
a saturated solution of ammonium acetate, acidified with HC 2 H 3 2 . 
Treat a portion of this solution with K 2 Cr0 4 solution. 

Yellow precipitate proves lead present. 
NOTE: Use a micro slide or test tube for this experiment. 

7. To the filtrate from step 6, add NH 4 OH until the solution is 
strongly ammoniacal. Filter, and reserve the filtrate for step 8. 
Dissolve the precipitate in the least volume of HC1 required. Place 
a drop of the solution on a piece of filter paper, and add a drop of con- 
centrated NH 4 OH. Treat the precipitate with a freshly prepared 
solution of sodium stannite. 

Black precipitate proves metallic bismuth. 

Add a drop of the acid solution to a piece of filter paper that has been 
previously treated with an equal volume of cinchonine reagent. 

Orange precipitate proves presence of bismuth. 

8. If the filtrate from step 7 has a deep blue color, the presence of 
copper is indicated. Neutralize a drop of the solution with acetic 
acid, and add K 4 Fe(CN) 6 solution. 

Red precipitate of Cu2Fe(CN) 6 indicates copper. 

NOTE: This test is much more sensitive than the blue complex obtained 
with ammonium hydroxide. 



GROUP II CATIONS 185 

Place a drop of the blue solution on a piece of filter paper and add 
a-benzoinoxime to the wet spot. Hold over an open bottle of NH 4 OH. 

Green spot confirms copper. 

9. If copper has been found to be present, carefully add KCN solu- 
tion to the remainder of the blue ammoniacal solution from step 7 until 
the color is completely discharged, and proceed as in stop 10. If no 
copper was found (colorless solution), make the solution faintly acid 
with H2S04, and proceed as in step 10. 

10. Saturate the colorless solution with H 2 S. 

Yellow precipitate of CdS indicates cadmium. 

NOTE: If a dark precipitate is obtained at this point, it indicates that 
the previous separations have been incomplete. In order to remove the 
traces of other Croup II A metals present, wash the precipitate thor- 
oughly, treat with H 2 SO 4 (1 part 6\V H 2 SO 4 to 4 parts H 2 O), and boil 
gently. Filter, make the filtrate ammoniacal, and filler again. Saturate 
the filtrate with H 2 S. If cadmium was present in the original precipitate, 
yellow CdS will form. 

Filter the precipitate, and digest with several drops of HNO 3 . Heat to 
expel excess HaS. Cool, and filter if necessary. Make the solution 
alkaline with NaOH, and add a crystal of thiosinamine. 

Yellow solution or precipitate proves cadmium. 

SUJKJKOUP B 

11. To the filtrate from stop 3, add IIC1 until the solution is dis- 
tinctly acid. 

NOTE: If none of this group is present, the precipitate will be white. 
Often a tannish precipitate is obtained in the absence of ( Jroup 1 1 B metals. 
This is due to the presence of minute amounts of CuS. CmS is slightly 
soluble in (NH 4 )sS*. 

Filter and discard the filtrate. Add 1 ml. of concentrated II Cl to the 
precipitate, and heat in boiling water for about 5 rnin., filter, and save 
the precipitate for step 14. 

NOTE: Arsenic sulfidc is not soluble in HC1 but may be converted to 
the volatile trichloride upon boiling with acid. Therefore, the tempera- 
ture in the preceding treatment must not be more than 100O. 

12. Evaporate the filtrate to about one-half the original volume to 
remove excess H 2 S. Place a drop of the concentrated solution on 
silver coin, and touch the drop with a piece of tin. 

Jet black spot, forming almost immediately, indicates 
antimony. 



186 QUALITATIVE ANALYSIS 

Dilute 2 drops of the solution with an equal volume of water. Add a 
small amount of solid NaN0 2 or KN0 2 . 

NOTE: The dilution should not be carried to the point of precipitation. 
If precipitation docs occur, add dilute HC1 until the precipitate just redis- 
solves. Nitrite is added at this point to oxidize the antimony to the 
pentavalent state. The previous treatment of the mixed sulfides with 
concentrated HC1 reduced the antimony to the trivalent state. The 
following test must be carried out with antimony in the oxidized state. 

Shake the tube and contents until no further evolution of gas takes 
place. Fill one of the depressions in a spot plate almost to the rim 
with rhodamine B solution. Add 2 drops of the oxidized antimony 
solution. 

Color change to violet indicates antimony. 

NOTE: It is best to perform this test in comparison with a control. Use 
a drop or two of pentavalent antimony solution in order to observe 
the proper color change. If either of the tests is ambiguous or incon- 
clusive, perform either the hydrolysis test or the "antimony cinnabar" 
test as given in the preliminary experiments. 

13. Treat the remainder of the solution with a 1- to 1.5-cm. strip of 
magnesium ribbon to reduce the stannic ion to the stannous condition, 
and filter rapidly. Divide the filtrate into two portions. 

NOTE: Use excess magnesium, if necessary. The reaction should be 
fairly vigorous, ft is important that all the metal be permitted to dis- 
solve before filtering. Add more acid, if necessary, to effect complete 
solution. 

Place one portion of the solution in a test tube and add HgCl 2 solution. 
White precipitate turning gray indicates tin. 

Place 1 or 2 drops of the second portion of the solution on filter paper, 
and treat with an equal volume of cacotheline solution. 

Violet color proves the presence of tin. 

14. Dissolve part of the precipitate of As 2 S 5 , obtained in step 11. in 
warm NH 4 OH to which have been added several drops of H 2 2 . 

NOTE: If no appreciable solution takes place, the residue is probably 
elementary sulfur and may be discarded. 

Place the solution in a crucible, add magnesia mixture, evaporate to 
dryness, and ignite. Allow the solid to cool, and add several drops of 
Bettendorf s reagent. 

Brown coloration, gradually deepening, indicates arsenic. 
The rest of the precipitate is dissolved in aqua regia and the solution 



GROUP II CATIONS 



187 



ANALYTICAL TABLE II 

ANALYSIS OF GROUP II 
Filtrate from Group I separation may contain Hg+ + , Pb+ + , Bi+++, Cu ++ , Cd+ 4 , 

As +++ , As +++++ , Sb +++ , Sb + + ++ +, Sn++, and Sn +4 ++ as well as the cations of 
Groups III, IV, and V. Adjust the acidity as directed, and saturate with H 2 S (1) : 
HgS (black), PbS (black), Bi 2 S 3 (brown-black), CuS (black) CdS (yellow), As 2 S.< 
(yellow), As 2 S & (yellow), Sb 2 S 3 (orange), Sb 2 S 6 (orange-), SnS (brown), SnS 2 
(yellow). Filter, acidify with HC1, boil until H 2 S is expelled, and reserve the fil- 
trate for subsequent group analyses (2). Extract precipita e with warm 
solution (3). 



Precipitate: HgS, PbS, BiA, CuS, CdS. 
Proceed as in Analytical Table III. 



Filtrate : AsSr, SbSr, SnS,- 
as in Analytical Table? IV. 



Proceed 



ANALYTICAL TABLE III 

ANALYSIS OF GROUP II A 
Heat residue from Table II with I1NO 3 to boiling and filter (4). 



Precipitate: HgS. 


Filtrate: Pb+^, Bi f< % Cu 4 % Cd+<. Boil with concentrated 


Dissolve por- 


H 2 SO 4 to fumes of SO,, cool and dilute with 1I 2 (). Filter (6). 


tion in aqua 








regia, expel ex- 


Precipitate :PbSO 4 


Filtrate : Bi^ M , Cu++, C^d -+. Make strongly 


cess reagent 


(white). Dis- 


ammoniac:) i and filter (7). 


and test for 


solve in 




Hg + + with 
SnCl 2 solution. 
Confirm with 
diphenylcar- 
bazidetest (6). 


NH 4 C 2 II 3 O 2 - 

HC 2 II 3 O a 
solution and test 
for Pb H ' with 
OO 4 " solution 


Precipitate: Hi (OFT), 

(white). Dissolve- 
in IK /I, reprecipi- 
tate on the filter 
paper with NH 4 OII 


Filtrate :Cu(NII 3 ) 4 +4 - 

(deep blue) and Cd- 
(NH S ) 4 * + (colorless). 
Neutralize a portion 
with HC 2 H 8 O 2 and 




(6). 


and test with fresh 


test for (>u^ with 






Na 2 Sn() 2 solution. 


K 4 Fe(CN) fi solution. 






Confirm with cin- 


(Confirm with -ben- 






chonine test (7). 


zoinoximo (8). If 








Cu ff present, treat 








another portion with 








KCN until colorless 








(9). Saturate with 








HaS to test for 








Cd f *. Confirm 








with thiosinamine 








test (10). 



evaporated to dryness. The residue is taken up in a small volume of 
water made slightly acid with acetic acid and treated with a small 
amount of solid sodium acetate and AgN() 3 solution added. 

Chocolate brown precipitate of Ag 3 AsO 4 proves arsenic. 

NOTE: The test should be carried out on a spot plate. The precipitate 
should form immediately. Formation of a brown precipitate after 



188 



QUALITATIVE ANALYSIS 



several minutes means that the silver nitrate has been converted to Ag 2 S 
by H 2 S fumes in the laboratory. 

ANALYTICAL TABLE IV 

ANALYSIS OF GROUP IIB 

Treat filtrate from Analytical Table II with dilute HC1 until distinctly acidic, 
and filter. Precipitate may contain As 2 S 6 , Sb 2 S 4 , SnS 2 . Heat the precipitate 
with coned. HC1, and filter (11). 



Precipitate : As 285. Dissolve a portion in 
warm NH 4 OH-H 2 O 2 solution. Treat 
with magnesia mixture, evaporate to 
dryness, and ignite. Test with Bet- 
tendorf's reagent. Dissolve another 
portion of the As 2 S 6 in aqua regia. 
Evaporate to dryness. Take up in 
H2O, make slightly acid with HC2H 3(1)2, 
and add AgNO 3 (14). 



Filtrate: Sb ++ +, Sn ++++ . Evaporate 
and test portion for Sb+ ++ with Sn-Ag 
couple. Confirm by oxidation to pen- 
tavalent state, and test with rhoda- 
mine B solution (12). Reduce 
remainder of solution with Mg, filter 
and test portion with HgCl 2 solution. 
Confirm Sn ++ with cacotheline test 
(13). 



GROUP III CATIONS 

ALUMINUM, Al 

Aluminum, a silvery, fairly soft metal, was first isolated in the early 
part of the nineteenth century, notwithstanding the fact that its com- 
pounds comprise a large part of the earth's crust. For a long time it 
was as expensive as platinum, but with the advent of the Hall electro- 
lytic process, the price quickly dropped. The metal is an excellent 
conductor of heat and electricity. It is very light and is used exten- 
sively in the manufacture of strong, lightweight alloys, cooking 
utensils, and automobile parts. Duralumin is an important alloy 
used in construction where lightness and strength are required. 
Aluminum is placed in Group 3 of the periodic table. 

When pure, aluminum is only very slightly oxidized in air. It is 
soluble in hot, concentrated sulfuric acid with the evolution of sulfur 
dioxide. The metal dissolves in hydrochloric acid with the evolution 
of hydrogen, as might be expected from the position of aluminum in 
the electromotive series. When treated with nitric acid, the metal 
becomes "passive" and so unreactive that the concentrated acid can be 
shipped in aluminum drums. Aluminum reacts with alkalis, liberating 
hydrogen and forming the amphoteric meta-aluminate ion, A1O 2 ~". 

The cation exists only in the trivalent form. Its salts are generally 
colorless. Many aluminum compounds hydrolyze in aqueous solution. 
The sulfide, AUSs, is unstable in solution, and the cation is therefore 
always precipitated as the hydroxide, A1(OH) 3 , in analytical proce- 
dures. The base is amphoteric, reacting with excess alkali to form the 



GROUP III CATIONS 189 

meta-aluminate ion. Treatment of a soluble aluminum salt with 
ammonium benzoate solution precipitates the hydroxide quantita- 
tively in filterable form through hydrolysis of the reagent. Ammonium 
hydroxide produces a gelatinous precipitate of aluminum hydroxide, 
which is insoluble in the presence of ammonium salts. 

Aluminon and alizarin S Blue form colored "lakes" with aluminum 
hydroxide, which is precipitated by ammonia. When the solid hydrox- 
ide is strongly heated with cobalt nitrate solution, Co(NC) 3 ) 2 , blue 
cobalt aluminate, CoAl 2 O 4 , Thdnard's Blue, is produced. 

PRELIMINARY EXPERIMENTS 

1. Add NH 4 OH to a test solution of A1+++. Repeat tho procedure, 
adding some solid NH 4 C1 to the solution before precipitation. 

2. To the test solution, add NaOH slowly until an excess has been 
added. Neutralize the alkaline solution with dilute HC1. Add a few 
more drops of acid. 

3. Dilute 1 ml. of A1+++ test solution with an equal volume of 
water. Add dilute NH 4 ()H carefully until a very slight precipitate 
forms. Neutralize with dilute HCaH 3 O2, add a small amount of solid 
NH 4 C1, and pour the solution with stirring into 1 ml. of ammonium 
benzoate solution. Heat on a steam bath for 4 to 5 min., filter, and 
wash the precipitate. 

4. Place a few shreds of moist asbestos fiber in the loop of a platinum 
wire, dip it into some of the A1(OH) 3 prepared previously, and heat to 
dryness in the flame. Touch the loop to some dilute Co(NO 3 )'2 
solution, and heat strongly in the flame. 

5. Dissolve a portion of the A1(OH) 3 in dilute HC1, add a few drops 
of ammonium acetate solution and some aluminon reagent. Make 
the solution alkaline with NH 4 OH. 

6. Repeat the foregoing procedure, using alizarin S Blue in place 
of the aluminon. 

IRON, FE 

Iron, one of the most widely used of all metals, has been known 
since prehistoric times. It is used primarily in the manufacture of a 
large number of steels and alloys. Iron falls in Group 8 of tho 
periodic table, along with cobalt and nickel. Since iron is well above 
hydrogen in the electromotive series, it dissolves in both hydrochloric 
acid and sulfuric acid with the evolution of hydrogen. The action 
of nitric acid on the metal depends upon the concentration of the acid. 
Dilute nitric acid is reduced to ammonium ion, NH 4 +, without the 
evolution of any gas, and the metal is oxidized to the ferrous state. 



190 QUALITATIVE ANALYSIS 

With more concentrated acid, the gas NO is produced, and the iron is 
converted to the trivalent form. Concentrated nitric acid, in the cold, 
makes iron "passive." 

In its compounds, iron exists chiefly in the di- and trivalent states. 
When iron is anodic, or rusting, it is believed that it exists in tetra- 
and pentavalent forms, but these compounds have no analytical 
importance and will not be considered. The trivalent form of the ion 
is the more stable, ferrous ion being easily oxidized in air to the ferric 
state. Ferrous compounds are pale green in solution; ferric salts are 
yellow. The hydroxides are not amphoteric. 

Alkalis and ammonium hydroxide precipitate white to green ferrous 
hydroxide, Fe(OH) 2 , from ferrous solutions. The precipitate is 
oxidized in air to brown ferric hydroxide, Fe(OH) 3 ; the latter may be 
obtained directly from ferric salts by the action of the same reagents. 
In the presence of ammonium salts, ferrous hydroxide is not precipi- 
tated by the action of ammonium hydroxide. 

Hydrogen sulfide does not precipitate an iron sulfide in acid solu- 
tion. In ammoniacal solution, black ferrous sulfide, FeS, and ferric 
sulfidg, Fe2S 3 , are precipitated from the corresponding ions. Both 
sulfidcs are readily soluble in hydrochloric acid, with the evolution 
of hydrogen sulfide and the formation of ferrous chloride, FeCl 2 . 
Ammonium benzoate, in neutral solution, precipitates the hydroxides 
of the cations through hydrolysis of the reagent. 

With alkali cyanides, ferrous ion forms yellow ferrous cyanide, 
Fe(CN) 2 , which reacts with an excess of reagent to form the soluble 
complex yellow ferrocyanide (hexacyanoferrite) ion, [Fe(CN) 6 ]-. 
This complex gives a white precipitate of ferrous ferrocyanide, 
Fc 2 [Fe(CN) 6 ], with ferrous ion; with ferric ion, it forms a deep blue pre- 
cipitate, Fe 4 [Fe(CN) 6 ] 3 , which is called Prussian Blue. This reaction 
furnishes an extremely sensitive test for the trivalent cation. The 
compound itself is used as a paint pigment. 

Ferric ion also forms a soluble complex with an excess of cyanide 
ion, the orange ferricyanide (hexacyanoferrate) ion, [Fe(CN) 6 ] s . 
With ferrous ion, this complex forms an insoluble blue compound, 
Turnbull's Blue, Fe 3 [Fe(CN) 6 ]2; with ferric ions it forms brown 
Fe[Fe(CN) 6 ]. The thiocyanate ion, CNS~, gives rise to an intensely 
red-colored solution with ferric ion; this is due to the formation of 
[FeCNS] ++ but the CNS~~ gives no reaction with ferrous ion. This is 
also a sensitive test for the ferric ion and permits its detection in the 
presence of ferrous ion. 

Alkali acetates, in an acetic acid solution of the trivalent cation, 
form red, un-ionized ferric acetate, Fe(C 2 H 3 02)3. Boiling the solution 



GROUP III CATIONS 191 

precipitates red basic ferric acetate. Phosphates precipitate yellow- 
ish-white ferric phosphate, FePO 4 , which is insoluble in acetic acid. 
If phosphates are present in a solution to be analyzed, they must be 
removed before testing for iron (see Systematic Analysis, p. 262). 

PRELIMINARY EXPERIMENTS 

1. Add NEUOH to portions of Fc++ and Fe+++ test solutions. 
Filter and expose a portion of Fe(OH) 2 to the air for several minutes. 

2. Treat some Fe++ test solution with dilute NaOH. To tho 
suspension add a few particles of solid Na 2 2 . 

3. Make separate portions of the two test solutions ammoniacal 
with NH 4 OH, and add some solid NH 4 C1. Saturate both solutions 
with H 2 S. Observe the behavior of the precipitate from the ferrous 
salt on exposure to air. Treat separate portions of the precipitates 
with dilute (1:9) HC1. 

4. Treat a portion of Fe +++ solution with some potassium ferro- 
cyanide solution. Repeat the test with Fc" 4 "*. Add K 3 Fo(CN) 6 to 
each of the iron test solutions. 

5. Add a few drops of KCNS or NH 4 CNS to a small volume of 
Fe ++ + solution. 

6. Carry out the reaction with ammonium borizoato, as given under 
Aluminum. 

7. Add some NH 4 C 2 H 3 O 2 solution to mi acetic acid solution of 
ferric salt. Heat the solution to boiling. 

CHROMIUM, Cr 

Chromium is placed in Group 6 of the periodic table. It is a 
silvery, brittle metal, soluble in acids with the evolution of hydrogen 
and soluble in excess alkali with the formation of tho chromite anion 
CrO;f. In nitric acid, chromium is insoluble, becoming "passive" 
like aluminum and iron. Large amounts of chromium are used in tho 
manufacture of stainless steels and in electroplating. 

In its compounds, chromium usually exhibits valences of 2, 3, and 6. 
The divalent chromous compounds are very unstable, being easily 
oxidized in air to the more stable chromic ion, Cr f++ . Chromous 
compounds are, therefore, not considered in analytical procedures. 
Chromic ion may also be oxidized by chemical oxidizing agents to the 
hexavalent chromate, CrOr, and dichromate, O 2 O 7 ", anions. 

Alkalis, ammonium hydroxide, ammonium sulfide, and ammonium 
carbonate precipitate gray-green to blue chromic hydroxide, Cr(OH) 3 , 
from chromic ion solutions. Chromic sulfide does not exist in aqueous 
solution. Ammonium benzoate quantitatively precipitates the hydrox- 



192 QUALITATIVE ANALYSIS 

ide from neutral solutions through hydrolysis of the reagent, as men- 
tioned previously. Chromic hydroxide is soluble in acids, with the 
formation of chromic salts. Since it is amphoteric, it also reacts with 
alkalis to form the soluble chromite ion, Cr0 2 ~. Chromic phosphate is 
greenish in color and is soluble in acetic and mineral acids. 

Alkaline solutions of chromite ion and chromic hydroxide are 
readily oxidized to chromate ion with sodium peroxide, sodium 
hypochlorite, bromine, or chlorine. Chromic compounds are also 
oxidized to chromates when fused with sodium carbonate and sodium 
peroxide. In acid solution, chromate ion is converted to dichromate 
ion without any valence change : 

Cr 2 0r + H 2 ^ 2HCr0 4 - 
H 2 + HCrOr ^ H 3 0+ + CrOr 

Dichromates may be obtained by strong oxidation of chromic ion in 
acid medium. 

Barium, lead, and silver chromates are distinctly colored. The 
lead salt, known as chrome yellow, is a valuable pigment in the paint 
industry. 

With cold sulfuric acid and hydrogen peroxide, the chromate ion 
gives a deep blue coloration of so-called perchromic acid, Cr0 5 , which 
is readily soluble in ether. This compound is unstable, but the color 
persists longer in the ether layer than in aqueous solution. Cold 
concentrated sulfuric acid forms red crystals of chromium trioxide, 
CrOs, with dichromates and chromates. This mixture of crystals 
and acid is the active agent in "cleaning solution." 

Ammoniacal solutions of chromate ion produce a blue-colored 
complex with benzidine; in acid solution, the same ion gives a violet 
color with diphenyl carbazide. These sensitive identification reactions 
also furnish an excellent example of the variation of color in coordina- 
tion complexes. 

PRELIMINARY EXPERIMENTS 

1. Add NH40H to Cr +++ test solution until an excess has been 
added. Repeat the test, first adding solid NH 4 C1 to the test solution. 
Saturate the ammoniacal solution with H 2 S. 

2. Precipitate Cr(OH) 3 with ammonium benzoate, as described 
previously under aluminum. Dissolve the precipitate in dilute HC1. 
Add NaOH slowly to a portion of the solution; then add an excess. 
Neutralize the alkaline solution with dilute HC1. To another portion 
of the alkaline solution obtained above, add Na 2 2 ; heat till reaction 



GROUP III CATIONS 193 

ceases. Acidify the solution with dilute HC1. Finally, make alkaline 
with NaOH. 

3. Neutralize a portion of the alkaline chromate solution from the 
preceding experiment, and then acidify with, acetic acid. Add a small 
amount of solid NaC 2 H 3 02, and then add BaCl 2 solution. Repeat, 
using concentrated AgN0 3 solution in place of the BaCl 2 . 

4. Neutralize a second portion of the chromate solution, acidify 
with acetic acid, and add a few drops of benzidine solution. 

5. Make a third portion of the chromate solution acid with dilute 
H 2 S04, and add an equal volume of diphenyl carbazide solution. 

6. Add some H 2 O 2 to a solution of chromate ion acidified with dilute 
H 2 SO 4 . Add some ether, and shake. (CAUTION) 

NICKEL, Ni 

Nickel is placed in Group 8 of the periodic table with cobalt and 
iron. It is a hard, silvery- white magnetic metal that may be converted 
to the passive state by treatment with concentrated nitric acid. In 
finely divided form, nickel is an excellent catalyst for the hydrogenation 
of fats, oils and many classes of organic compounds. The metal is also 
used extensively in the making of coins and alloys and in electroplating. 
Some of the important nickel alloys are nickel steel, Nichrome, and 
Monel metal. 

Although nickel is above hydrogen in the electromotive series, it is 
difficultly soluble in hydrochloric acid and sulfuric acid but is readily 
soluble in moderately concentrated nitric acid. The nickel cation 
usually exhibits a valence of 2. In solution, it is pale green; the 
anhydrous salts, however, are yellow. Alkalis precipitate nickelous 
hydroxide, Ni(OH) 2 , which is insoluble in excess reagent but is soluble 
in acids. Ammonium hydroxide precipitates, from nickel sulfate 
solution, the green basic salt, Ni 2 S04(OH) 2 , which is soluble in excess 
reagent with the formation of the complex ion, [Ni(NH 3 ) 6 ] ++ . If 
ammonium salts are present in the ammonium hydroxide, no precipi- 
tate is obtained, the deep blue complex ion being formed immediately. 
Alkali carbonates precipitate apple-green nickelous carbonate, NiCOs. 
The corresponding compound produced with ammonium carbonate is 
soluble in excess reagent with the formation of the complex hexam- 
mino salt. 

In acid solution, the sulfide ion concentration of hydrogen sulfide 
is not sufficient to precipitate nickel sulfide. In neutral or ammoniacal 
solution, black nickel sulfide, NiS, is readily formed. The sulfide is 
very slowly soluble in 1 :9 HC1, slowly soluble in the concentrated acid, 
and readily soluble in nitric acid and aqua regia. 



194 QUALITATIVE ANALYSIS 

Solutions of nickelous ion give a brilliant red crystalline precipitate 
with dimethyl glyoxime when made faintly ammoniacal. The borax 
bead obtained with nickel ion is reddish brown in color in the oxidizing 
flame. 

PRELIMINARY EXPERIMENTS 

1. Add dilute NH 4 OH drop wise to Ni++ test solution, until an 
excess has been added. Repeat the foregoing, adding solid NH 4 C1 
to the test solution. Saturate the ammoniacal solution with H 2 S. 
Filter and wash the precipitate. 

2. Treat a portion of the sulfide with 1:9 HC1. Treat separate 
portions of the precipitate with HN0 3 and aqua regia. 

3. Make a portion of the test solution ammoniacal, and place a drop 
on a spot plate. Treat with an equal volume of dimethyl glyoxime. 
Acidify with HC1; make ammoniacal again. 

4. Fuse some borax in the loop of a platinum wire. Cool the bead, 
moisten with water, and touch it to the surface of a dilute Ni ++ solu- 
tion. Heat strongly in the oxidizing flame. 

COBALT, Co 

Cobalt, a steel-gray magnetic metal, was first isolated in the early 
nineteenth century. It is placed in Group 8 of the periodic table 
and is just above nickel in the electromotive series. Cobalt forms 
many alloys, among them the extremely hard stellite and carboloy. It 
is used to impart a blue color to glass and enamel and also as an oxida- 
tion accelerator for drying oils. The metal is readily soluble in dilute 
nitric acid and in hot, dilute hydrochloric and sulfuric acids. With 
concentrated solutions of nitric acid, cobalt becomes passive. 

The cobalt cation generally exhibits a valence of 2. Simple 
trivalent cobaltic salts are unknown and exist only in complex com- 
pounds like the cobaltinitrites (nitritocobaltates), cobalticyanides 
(cyanocobaltates), and complex ammines. In contrast to the pink 
color of the hydrated ion, anhydrous cobaltous salts are blue. Use 
is made of this sharp color difference in the manufacture of simple 
weather indicators. 

Alkalis precipitate a blue, basic cobalt salt, which decomposes on 
heating to pink cobaltous hydroxide, Co(OH) 2 . This is oxidized in air 
to the brown cobaltic hydroxide, Co(OH) 3 . Ammonium hydroxide 
also precipitates a blue basic salt, which is soluble in ammonium 
chloride, forming a brown solution that is gradually oxidized in air to 
the very stable chlorpentamminocobaltic chloride, [Co(NH 3 )5Cl]Cl 2 . 



GROUP III CATIONS 195 

Alkali carbonates precipitate a reddish colored basic cobalt carbonate 
that is soluble in ammonium carbonate and ammonium chloride. 

High concentrations of sulfide ion, such as are obtained in neutral 
or ammoniacal solution, precipitate black cobalt sulfide, CoS, which is 
very slowly soluble in 1:9 HC1 and in acetic acid. The sulfide is 
readily soluble in nitric acid and aqua regia with the separation of 
elementary sulfur. 

When concentrated solutions of cobaltous ion are treated with 
gfbluble nitrites and acetic acid, the complex cobaltinitrite (hexa- 
nitritocobaltate) ion, [Co(N0 2 ) 6 ]=, is produced. If potassium nitrite is 
used in this preparation, a crystalline yellow precipitate of potassium 
cobaltinitrite is obtained. When concentrated solutions of ammonium 
thiocyanatc, NH 4 CNS, are added to cobaltous salts, a blue solution 
of the complex cobaltithiocyanate (tetrathiocyanatocobaltate) ion, 
[Co(CNS)]4 = , is obtained. The complex is very soluble in amyl 
alcohol or a mixture of amyl alcohol and ether, which intensifies the 
color. The addition of acetone to the solution also intensifies the 
color. Ferric ion, in small amounts, interferes with this test, but the 
interference may be minimized by the addition of tartaric acid to 
the solution before testing. 

A purple-red precipitate is obtained when a freshly prepared solu- 
tion of a-nitroso-/3-naphthol is added to an acetic acid solution of cobalt 
ion. The colored precipitate is insoluble in cold, dilute acids. The 
borax bead coloration in either the oxidizing or reducing flame is blue. 
This is an extremely sensitive and conclusive test. 

PRELIMINARY EXPERIMENTS 

1. Pass H 2 S into an ammoniacal solution of cobalt ion, containing 
some ammonium chloride. Treat a portion of the precipitate with a 
few drops of 1:9 HC1. Add aqua regia to a second portion of the 
sulfide. 

2. Prepare a borax bead as described previously, and touch it to a 
little of the cobalt sulfide. Heat strongly in the oxidizing flame. 

3. Acidify a portion of a concentrated Co++ solution with dilute 
HC2H 3 2 , saturate the solution with solid KC1, and then add an excess 
of KN0 2 solution. 

4. Make a portion of the test solution acid with acetic acid. Place 
a drop of the solution on a spot plate, and add an equal volume of 
a-nitroso-/3-naphthol solution. 

5. Acidify 1 ml. of test solution with acetic acid; treat with an equal 
volume of NH 4 CNS solution in acetone. 



196 QUALITATIVE ANALYSIS 

MANGANESE, Mn 

Manganese is a grayish- white metal that is very easily oxidized in 
moist air. It was discovered in 1774 by the Swedish chemist Scheele. 
It is a transition element, falling in Group 7 of the periodic table. 
The metal is used in the manufacture of certain types of steel, in dry 
cells in the form of manganese dioxide, Mn0 2 , as a dryer for paints and 
oils, and as a coloring agent for glass and ceramics. 

Manganese is a fairly active element, being in the upper part of 
the electromotive series, just below aluminum. It reacts with warm 
water, evolving hydrogen and forming a precipitate of manganous 
hydroxide, Mn(OH)2. It is readily attacked by dilute mineral acids, 
forming manganous ions. 

The element exists in a number of valence states 2, 3, 4, 6, and 7. 
It exhibits, also, a gradation of properties, ranging from a basic cation 
in Mn 4 "" 1 " to the strongly acidic anion, Mn04~, where manganese has 
a valence of 7. The divalent manganese ion is most generally used in 
analytical procedures. It forms pink compounds, both in solution and 
in crystalline form, but is colorless when anhydrous. Alkalis precipi- 
tate white manganous hydroxide from manganous ion test solution; 
this turns brown in air due to oxidation to MnO(OH) 2 . Manganous 
hydroxide is insoluble in excess reagent but is soluble in ammonium 
salts before oxidation. Ammonium hydroxide precipitates the base 
incompletely, and not at all in the presence of ammonium salts. 

Sulfide ion precipitates pink manganous sulfide, MnS, from ammo- 
niacal solution. The sulfide is soluble in hydrochloric acid. By the 
action of hydrogen sulfide on a hot ammoniacal solution of manganous 
ion, green MnS may be precipitated. The same compound may be 
obtained by the reduction of manganate, Mn0 4 =a , and permanganate 
ions in alkaline solution, in the presence of sulfide ion. All the 
higher valence forms are reduced to manganous ion by sulfide 
ion. Sulfites reduce manganate and permanganate ion to manganese 
dioxide, Mn0 2 , in neutral solution and to manganous ion in acid 
solution. 

Oxidizing agents, like sodium peroxide and sodium hypobromite, 
convert manganous ion to manganese dioxide. Stronger oxidizing 
agents, like lead dioxide, PbO2, in acid solution or ammonium per- 
sulfate in the presence of silver nitrate or sodium bismuthate, NaBiOs, 
oxidize manganous and manganic ions to the violet-colored perman- 
ganate ion, MnO 4 ~. Halogen acids function as reducing agents when 
treated with the higher oxidized forms, Mn(>2, MnC^, and MnOr", 
the free halogen being produced. Oxidative fusion of manganous 



GROUP III CATIONS 197 

hydroxide with sodium carbonate produces a green bead of sodium 
manganate. 

PRELIMINARY EXPERIMENTS 

1. Treat Mn++ test solution with NH 4 OH. Add solid NH 4 C1 
to the suspension. Add a mixture of NEUOH and NH 4 C1 to another 
portion of the test solution. 

2. Saturate an ammoniacal solution of Mn + + with H 2 S. Boil the 
suspended solid. Filter and ascertain the solubility of the sulfide in 
1:9 HC1. 

3. Add NEUOH to a portion of the test solution. Treat the pre- 
cipitate with a small amount of solid Na20 2 . Acidify a portion of the 
test solution with dilute HN0 3 , add a crystal of KC1O 3 , and heat to 
boiling. Cool, and repeat the treatment. Save the precipitate for 
experiments 4 and 5. 

4. To a portion of the washed precipitate from experiment 3, add 
dilute HNO 3 and a few drops of H 2 O 2 . Boil to expel excess peroxide, 
cool, and add a few crystals of NaBi() 3 . Shako, and allow the solid to 
settle. 

5. Dissolve another portion of the washed precipitate from experi- 
ment 3 in dilute HNO 3 , add 1 drop of AgNO 3 solution and a few crystals 
of ammonium persulfate, and warm. 

6. Fuse a small amount of Na 2 CO 3 in the loop of a platinum wire, 
and dip it into some Mn(OH) 2 . Heat in the flame, touch the bead to 
some powdered KC1O 3 , and heat again. 

ZINC, Zn 

Zinc is a bluish-white low-melting metal that is placed in Group 2 
of the periodic table. It finds wide use in commerce as a protective 
coating for iron (galvanizing), as a constituent of paint pigments, as a 
component of alloys, notably brass, in dry cells and in pharmaceutical 
preparations. 

The metal is above hydrogen in the electromotive series and dis- 
solves in hydrochloric and sulfuric acids. It is, however, a strong 
reducing agent and gives nitrogen dioxide, nitric oxide, and ammonia 
with varying concentrations of nitric acid. Like aluminum, it is solu- 
ble in alkalis with the evolution of hydrogen and the formation of the 
amphoteric zincate ion, ZnOr. 

The zinc cation exists only in a valence of 2. Alkali hydroxides 
precipitate white zinc hydroxide, Zn(OH) 2 , which is soluble in excess 
reagent with the formation of the zincate ion, ZnO-f. In the absence 
of ammonium salts, ammonium 'hydroxide precipitates zinc hydroxide, 



198 QUALITATIVE ANALYSIS 

which is soluble in excess reagent with the formation of the colorless 
hexammino zinc ion, [Zn(NH 8 )6] 4 " f ". Ammonium sulfide precipitates 
white zinc sulfide, ZnS, from alkaline, neutral, or faintly acid solutions 
of zinc ion. Hydrogen sulfide will precipitate zinc sulfide from neutral 
solution or from a strongly buffered acetic acid solution. When zinc 
ion is added to a mixture of diphenylamine acetate and potassium 
ferricyanide, it catalyzes the oxidation of the amine; dark-colored 
products result. Pyridine and ammonium thiocyanate produce a 
heavy white precipitate with Zn ion; strong heating of a mixture of the 
ion with potassium cobalticyanide and potassium chlorate or with 
dilute cobalt nitrate solutions produces a green residue (Rinman's 
Green). 

PRELIMINARY EXPERIMENTS 

1. Pass H 2 S into an ammoniacal solution of Zn ++ test solution. 
Neutralize a portion of the test solution with NH 4 OII; then make just 
acid with acetic acid. Add some solid NH 4 C 2 H 3 2 , and saturate the 
solution with H 2 S. 

2. Add NH 4 OH cautiously to the test solution until an excess has 
been added. Repeat, first adding an equal volume of NH 4 C1 solution 
to the Zn 4 " 4 " test solution. 

3. Carefully add NaOH to the test solution until an excess has been 
added. Neutralize the solution with dilute HC1. 

4. Make a portion of the test solution faintly acid with acetic 
acid. Add 1 drop of diphenylamine acetate solution (1 g. of the amine 
in 100 ml. of glacial acetic acid) and 1 ml. of 5% K 3 Fe(CN) 6 solution. 
A greenish black turbidity is indicative of the presence of zinc. 

5. To a portion of the neutral test solution, add 1 drop of pyridine 
and several drops of NH 4 CNS solution. 

6. Place a drop of the neutral test solution on a piece of filter paper 
that has been treated with KC1O 3 and K 3 Co(CN) 6 reagent. Roll the 
paper into a small ball, and heat strongly in the loop of a platinum wire 
or in a crucible. 

(Alternate procedure) Place a few shreds of moist asbestos fiber in the 
loop of a platinum wire, dip it into some Zn(OH) 2 , and heat to dry ness 
in the flame. Touch to some dilute Co(N0 3 ) 2 solution, and heat 
strongly in the flame. 

ANALYSIS OF GROUP III CATIONS 

OUTLINE OF SEPARATIONS 

The cations of Group III are precipitated from an ammoniacal 
solution containing ammonium chloride and ammonium sulfide, the 



GROUP III CATIONS 199 

latter resulting from the neutralization of the base by hydrogen sulfide. 
Co ++ , Ni ++ , Fe+ + , Mn+ + , and Zn++ are precipitated as sulfides; A1+++ 
and Cr +++ are obtained as the hydroxides. 

The entire precipitate is redissolved in aqua regia and separated 
into two subgroups. The A subgroup, Al, Fe, and Cr, is precipitated 
as hydroxide in easily filterable form by the use of ammonium benzoate 
in neutral solution, leaving subgroup B in the filtrate. Sufficient 
hydroxyl ion for this precipitation and separation is furnished by 
hydrolysis of the ammonium benzoate. 

The hydroxides of subgroup A are redissolved, and on boiling with 
sodium peroxide and sodium hydroxide, Fe is reprecipitated as the 
hydroxide, leaving Al in the form of the soluble aluminate, while Cr is 
oxidized to CrC^. Aluminum is then separated by precipitation in a 
weakly alkaline medium. 

Subgroup B ions are reprecipitated as the sulfides ; NiS and CoS are 
separated from MnS and ZnS by their insolubility in 1:9 HC1. Man- 
ganese is precipitated from the solution as MnOu and thus separated 
from zinc. 

ANALYTICAL PROCEDURE 

The unknown solution may be the H 2 S-free slightly acid filtrate 
from the Group II separation, or it may be a mixture containing the 
csttions of this group only. 

1. Make the volume of solution up to 4 ml. with distilled water, add 
approximately 0.25 g. of solid NH 4 C1, and stir until it is completely 
dissolved. Add NHUOH until the solution is slightly alkaline to litmus; 
warm and saturate with HsS. When precipitation is complete, filter 
and wash the precipitate carefully with distilled water. If the filtrate 
is to be used for further analysis, boil it with a few drops of glacial 
acetic acid to expel H 2 S, filter if necessary, and reserve for Groups IV 
and V. 

2. Redissolve the precipitate in the least volume of aqua regia 
required, evaporate almost to dryness, dilute with 1 ml. of water, and 
filter if necessary. Add NH 4 OH to the filtrate until a very slight pre- 
cipitate forms. Clear the solution with acetic acid, add a few crystals 
of NH 4 C1 and 1 ml. of ammonium benzoate solution. Dilute to 10 ml. 
with distilled water, and heat on a steam bath for 4 to 5 min. Cool, 
filter, and wash the precipitate several times. Combine the filtrate 
and washings, and reserve for Group IILB analysis (step 8). 

SUBGROUP A 

In addition to the hydroxides of Al, Fe, and Cr, the precipitate also 
contains a small amount of benzoic acid. 



200 QUALITATIVE ANALYSIS 

3. Boil the precipitate with about 0.5 ml of dilute HC1. Evapo- 
rate to a small volume, add 1 ml. of water, and filter. Discard the 
precipitate of benzole acid. 

4. Make the solution slightly alkaline with dilute NaOH, and add a 
small amount of Na 2 O 2 . Boil gently for about 2 min., cautiously 
adding small amounts of Na 2 02 from time to time, with stirring, until 
an excess has been added. Cool the solution, add several drops of 
water, filter, and wash the precipitate with cold water. Combine the 
filtrate and washings and save for the Al and Cr tests in step 6. 

NOTE: Care should be taken not to add the peroxide too rapidly 
because of the violent reaction of Na2C>2 and water. Sodium peroxide 
should be transferred in small quantities from the container to a watch 
glass and never onto filter paper. 

5. Add just enough dilute HC1 to the precipitate to dissolve it 
completely, and dilute the solution with 3 to 4 drops of water. Place 
1 drop of the solution on a piece of filter paper, and add an equal volume 
of K 4 Fe(CN) 6 solution. 

Deep blue precipitate of Prussian Blue indicates ferric ion. 
Place 2 drops of the solution on a spot plate, and add 1 drop of KCNS 
solution. 

Intense red coloration of (FeCNS)++ proves ferric ion. 

6. Make the filtrate from step 4 slightly acid with dilute HC1. 

NOTE: The acid should be added cautiously while the solution is being 
cooled under running water to prevent reduction of Cr04~ to Cr ++ +. 

Add a small amount of solid NH 4 C1; then make alkaline with NEUOH, 
and heat to boiling. 

NOTE: The NFUOH should bo added to the solution until, after blow- 
ing away the vapors of Nils, the solution has a distinctly anunoniacal 
odor. 

Gelatinous precipitate indicates aluminum. 

Filter, and save the filtrate for step 7. Dissolve the precipitate in 
dilute HC1. Place a few drops of the solution in a test tube, add 1 
drop of alizarin S Blue, and make slightly ammoniacal. 

Blue "lake" indicates presence of aluminum. 

Place the remainder of the acid solution in a test tube, add 2 drops of 
aluminon reagent, and make the solution slightly ammoniacal. 

Red "lake" confirms the presence of aluminum. 

NOTE: If chromium is present, the filtrate from the aluminum separa- 
tion will be yellow. 



GROUP III CATIONS 201 

7. Evaporate the solution from step 6 to a small volume, and make 
it slightly acid with dilute H 2 S0 4 . Place 1 drop of the solution on a 
spot plate, and add an equal volume of diphenyl carbazide solution. 

Violet-blue coloration indicates chromium. 

Place the remainder of the solution in a test tube, and add an equal 
volume of ether. Add a few drops of H 2 0o, and mix thoroughly by 
shaking. 

Blue color in upper (ether) layer proves chromium. 

NOTE: Since ether is highly inflammable, it should be kept away from 
open flames. If the tube has been stoppered before shaking, care should 
be exercised in removing the stopper so that the volatile contents are 
not splashed about and into the face. 

SUBGROUP B 

The filtrate from step 2 contains the cations Ni '+, Co ++ , Mn ++ , and 
Zn +4 ~ and some ammonium benzoate. 

8. Add sufficient HC1 to render the solution distinctly acid, and 
filter off the precipitated benzoic acid. MrJke the filtrate slightly 
ammoniacal, add a few crystals of NH 4 C1, heat to boiling, and pass in 
BUS until precipitation is complete. Filter, wash the precipitate, and 
discard the filtrate. 

9. Place the precipitate in a crucible, add 1 ml. of 1:9 HC1, and 
allow the solid to digest in the cold for 8 to 10 min., stirring frequently. 
Filter, and reserve the filtrate for step 12. 

NOTE: Under these conditions, MnS and ZnS arc dissolved completely, 
leaving a black residue of NiS and OoS. If the digestion is permitted to 
proceed too long, small amounts of nickel and cobalt sulfidcs will be 
dissolved. * 

10. Dissolve the black precipitate in aqua regia, evaporate almost 
to dryness, take up the residue with 1 ml. of water, filter, and discard 
the precipitate. Place a drop of the solution on a spot plate, add dilute 
NH 4 OH until the solution is basic, and then add a drop of dimethyl 
glyoxime solution. 

Red precipitate proves presence of nickel. 

11. Place a few drops of the solution from step 10 in a test tube, 
and make distinctly acid with acetic acid. Add 2 drops of a freshly 
prepared solution of a-nitroso-0-naphthol. 

Intense red precipitate proves cobalt. 



202 



QUALITATIVE ANALYSIS 



To the remainder of the preceding solution add a solution of NH 4 CNS 
in acetone. 

Blue color proves cobalt. 

12. Make the filtrate from step 9 alkaline with NaOH, add a few 
granules of Na 2 2 , and heat gently for about 1 min. Repeat the treat- 
ment several times. Dilute the solution with a few drops of water, 
filter, and wash the precipitate with cold water. Combine the filtrates, 
and reserve for step 14. 

NOTE: The precipitate is hydrated Mn0 2 ; the solution contains Zn0 2 ~ 
ion. 

13. Wash part of the precipitate thoroughly with water and then 
dissolve it in hot dilute HNO 3 . Add a few drops of H 2 O 2 . Boil to 
decompose excess peroxide, cool, and add a little solid NaBi0 3 . Shake 
and permit the solid to settle. 

Purple coloration of Mn04~ proves manganese. 

14. Make a portion of the zinc filtrate from Step 12 just acid with 
acetic acid, add some solid NH 4 C 2 H 3 O 2 and saturate with H 2 S. 

White precipitate of ZnS indicates zinc. 

Make another portion of the alkaline solution faintly acid with acetic 
acid. Add 1 drop of diphenylamine acetate solution and 1 ml. of 
5% K,Fc(CN) e . 

Greenish-black turbidity proves the presence of zinc. 

ANALYTICAL TABLE V 

ANALYSIS OF GROUP III 

The filtrate from Analytical Table II (separation of Group II) may contain 
Ni++, Co ++ , Fe ++ , Fe +++ , Mn ++ , A1+++, Cr +++ , and Zn ++ , as well as the cations 
of Groups IV and V. Make the filtrate ammoniacal, add solid NH^Cl, and satu- 
rate with H 2 S. Filter, boil the filtrate with glacial HC 2 II 3 2 , and reserve for the 
Groups IV and V analysis [Analytical Tables VIII and IX (1)]. Dissolve the 
precipitate in aqua regia, evaporate almost to dryness, take up in H 2 0, and filter 
if necessary. Add NH 4 OH until a slight precipitate forms, clear with HC 2 H 3 O 2 , 
add solid NH 4 Cl and ammonium benzoate solution. Dilute with H 2 O to 10 ml., 
heat, cool, and filter (2). 



Precipitate: A1(OH) 3 (white), Fe(OH) 3 
(red-brown), Cr(OH)a (gray-green). 
(Group III A.) Proceed as in Ana- 
lytical Table VI. 



Filtrate: Ni+ + , Co + +, Mn++, Zn+ + . 
(Group lllB.) Proceed as in Ana- 
lytical Table VII. 



GROUP IV CATIONS 



203 



ANALYTICAL TABLE VI 

ANALYSIS OF GROUP III A 

Boil the precipitate from Analytical Table V with dilute HC1, treat with H 2 O, 
filter, and discard the precipitate of benzoic acid (3). Make the solution alkaline 
with NaOH, add excess Na 2 02 in small portions, and boil for several minutes. 
Cool, dilute with H2O, and filter. Wash the precipitate, and combine the filtrate 
and washings (4). 



Precipitate: Fe(OH) 3 . 
Dissolve hi dilute HC1, 
and test portion with 
K 4 Fe(CN) 6 solution. 
Confirm with KCNS 
solution (5). 



Filtrate: A1O 2 -, CrO 4 " (yellow solution if CrO 4 " 
present). Acidify with dilute HC1, add solid NH 4 C1, 
make ammoniacal, boil, and filter (6). 



Precipitate: A1(OH) 3 . 
Dissolve in dilute HC1, 
and test portion with 
alizarin S in ammoniacal 
solution. Confirm with 
aluminon reagent (6). 



Filtrate : CrO 4 ". Evapo- 
rate and acidify with 
dilute H 2 SO 4 . Test with 
H 2 O 2 and ether, and con- 
firm with diphenyl car- 
bazide solution (7). 



ANALYTICAL TABLE VII 

ANALYSIS OP GUOUP lllB 

Acidify the filtrate from Analytical Table V with HC1, and discard the precipitate 
of benzoic acid. Make the filtrate ammoniacal, add sulid NH 4 C1, boil, and saturate 
with H 2 S: NiS (black), CoS (black), MnS (pink), ZnS (white). Filter, wash the 
precipitate, and discard the filtrate (8). Treat the precipitate with l:9HClfor 
several minutes, and filter (9). 



Precipitate: NiS, CoS. 
Dissolve in aqua regia, 
evaporate almost to 
dryness, add H 2 O, fil- 
ter, and discard the 
precipitate. Tost one 
portion for Ni ++ with 
dimethyl glyoxime(lO). 
Test for Co ++ with 
a-nitroso-0-naphthol, 
and confirm with 
NH 4 CNS-acetone test 
(11). 



Filtrate: Mn++, Zn++. Make alkaline with NaOH, 
add Na 2 O 2? and heat. Cool, and repeat treatment 
several times. Dilute with H 2 O, filter, wash pre- 
cipitate, and combine washings and filtrate (12). 



Precipitate: MnO 2 
(hydrated, brown). Dis- 
solve portion of precipi- 
tate in hot HNO 3 -H 2 O 2 
mixture, boil, and test 
with NaBiQa (13). 



Filtrate: ZnO 2 -. Make a 
portion acid with 
HC s H,O, y add solid 
NH 4 C 2 H 3 O 2 , and saturate 
with H 2 S. Confirm with 
the diphenylamine 
acetate and K 3 Fe(CN) 6 
test (14). 



GROUP IV CATIONS 

CALCIUM, Ca 

Calcium is a silvery-white metal that tarnishes rapidly in air. Since 
it falls in Group 2 of the periodic table, its hydroxide is a strong base, 
although it is not very soluble in water. Calcium is divalent in all its 
compounds. 



204 QUALITATIVE ANALYSIS 

Compounds of calcium are widely distributed over the earth's crust. 
It occurs chiefly as the carbonate (marble, calcite, chalk, limestone) 
and as the sulfate (gypsum). It finds application in industry as 
plaster, [calcium hydroxide, Ca(OH) 2 ], plaster of Paris (calcium sulfate 
hemihydrate, CaSO^J^H^O), and in building materials. 

Ammonium carbonate and ammonium hydroxide precipitate the 
carbonate almost completely from calcium ion solutions. Calcium 
carbonate is insoluble in water, but it is appreciably soluble in the 
ammonium salts of strong acids (ammonium chloride, ammonium 
sulfate). Chromate ion precipitates yellow calcium chromate, 
CaCr04, from concentrated solutions of calcium ion. The precipitate 
is soluble in water and acetic acid. White calcium oxalate, CaC 2 4 , is 
precipitated by the action of oxalate ion on soluble calcium salts. It is 
Insoluble in water and acetic acid but is readily soluble in mineral 
acids. Calcium sulfate is only slightly soluble in water. 

Calcium compounds color the flame brick-red. With the hand 
spectroscope, the chief lines observed are a yellowish green to one side 
of the sodium line and an orange-yellow double lino on the other side. 

PRELIMINARY EXPERIMENTS 

1. Add ammonium carbonate solution to a portion of Ca ++ test 
solution. Treat the precipitate with acetic acid. 

2. To the tost solution, add some K^CrC^ solution. Treat some of 
the solution with two or three times its volume of ethyl alcohol, noting 
the volume added. Add acetic acid to the precipitate. 

3. Add dilute H^SCh to the test solution. 

4. Treat a portion of the test solution with ammonium oxalate 
solution. Test the solubility of the precipitate in acetic and mineral 
acids. 

5. Perform a flame test on a calcium salt. l Observe the flame both 
with the naked eye and with the hand spectroscope. 

STRONTIUM, Sr 

Strontium is very similar to calcium in many ways. The pure 
metal is generally described as silver-white, although some observers 
claim that it is brass-yellow. It is believed that this tint is due to 
impurities. Strontium falls in the same group of the periodic table as 
calcium. 

Strontium ion, divalent in all its compounds, is similar to calcium 
in its reactions with ammonium carbonate and ammonium oxalate. 
The chromate, SrCrO 4 , is less soluble than calcium chromate but 

1 See Flame Tests, Blowpipo. Analysis, p. 220. 



GROUP IV CATIONS 205 

much more soluble than barium chromate. The same relationship 
exists among the sulfates of the three cations of this analytical group. 
Strontium compounds color the flame crimson. This fact is used 
in the manufacture of fireworks and flares containing strontium salts. 
Viewed in the hand spectroscope, the spectrum of strontium shows a 
number of lines in the red and orange-yellow portions and one line in 
the blue region. 

PRELIMINARY EXPERIMENTS 

1. Treat a portion of Sr ++ test solution with (NH 4 ) 2 C() 3 solution. 
Add acetic acid to the precipitate. 

2. Add K 2 Cr()4 solution to the test solution. Treat some of the 
solution with ethyl alcohol, noting the comparative volume added. 
Treat the precipitate with acetic acid. 

3. Add dilute H 2 S0 4 to a portion of Sr+ + test solution. 

4. To a portion of test solution, add (NH 4 ) 2 C 2 4 solution. Test 
the solubility of the precipitate in acetic and mineral acids. 

5. Perform a flame test on a strontium salt. Observe the colored 
flame with the naked eye and with a hand spectroscope. 

BARIUM, Ba 

Barium is usually described as a silver-white metal and is placed in 
Group 2 of the periodic table. It resembles both calcium and stron- 
tium in its chemical properties. Soluble barium salts are extremely 
poisonous. Barium sulfide, B;iS, is soluble in water and is often used 
as a depilatory. 

Barium carbonate, BaC0 3 , exhibits chemical properties similar to 
those of calcium carbonate. Yellow barium chromate, BaCrO 4 , is pre- 
cipitated from neutral solution by alkali chromates. It is insoluble 
in acetic acid. Oxalate ion precipitates white barium oxalate, BaC 2 04, 
which is insoluble in water, although it is more soluble than the cor- 
responding calcium and strontium salts. It is readily soluble in hot 
dilute acetic acid. Sulfate ion precipitates white barium sulfate, 
BaS0 4 , which is very insoluble in water and hot concentrated mineral 
acids. 

The barium flame is yellowish green. In the hand spectroscope, 
there can be seen numerous deep green and some weak green lines in 
the spectrum. 

PRELIMINARY EXPERIMENTS 

1. Add (NH 4 )2C0 3 to Ba+ + test solution. Treat the precipitate 
with acetic acid. 



206 QUALITATIVE ANALYSIS 

2. Treat a neutral solution of Ba ++ with K 2 Cr04 solution. Test the 
solubility of the precipitate in acetic acid. 

3. To the test solution, add (NH 4 )2C 2 O4 solution. 

4. Treat a portion of the test solution with dilute H^SCh. Test the 
solubility of the precipitate in hot water and mineral acids. 

5. Perform a flame test on Ba ++ test solution. Observe with the 
naked eye and with the hand spectroscope. 

6. (Optional) Test lithopone pigment for barium as follows: Mix 
powdered charcoal with the lithopone, and hold in the flame in the loop 
of a platinum wire. Moisten with dilute HC1, and hold in the flame 
again. 

ANALYSIS OF GROUP IV CATIONS 
OUTLINE OF SEPARATIONS 

The metals of Groups IV and V are converted to the chlorides, and 
any excess ammonium salts present are removed. The cations of this 
group are then precipitated individually from solution, no group 
reagent |?eing employed. 

Barium ion is precipitated as the chromate, strontium as the sulfate, 
and calcium as the oxalate. The remaining solution is reserved for the 
analysis of Group V. 

ANALYTICAL PROCEDURE 

1. Evaporate the filtrate from previous group separations to a small 
volume, filter, and discard any precipitate. Add concentrated HC1 
(1 ml. /ml. of solution), and again evaporate to a very small volume but 
not to dryness. Again add the same volume of concentrated HC1, and 
evaporate just to dryness. DO NOT IGNITE. Add about 1 ml. of 
distilled water, and again evaporate cautiously just to dryness. Dis- 
solve the residue in about 1 ml. of 0.2^ HC1. (All the material should 
dissolve.) 

2. Heat the solution to boiling, and add an 0.5^V solution of 
(NH4)2Cr04, drop by drop, until precipitation is complete. 

Yellow precipitate indicates barium. 

NOTE: The solution should be kept warm and the (NH^CHX added 
drop wise until further addition produces no turbidity in the clear super- 
natant liquid. The coloration of the solution at this point should be 
intensely yellow. 

Filter, and save the filtrate for step 3. Confirm the presence of 
barium in the precipitate by a flame test. 

Yellowish-green flame confirms barium. 



GROUP V CATIONS 



207 



3. Heat the filtrate from step 2 to 70 to 80C., and add a solution of 
0.5AT (NH 4 )2SO 4 drop wise until no further precipitation occurs. 
Filter, and reserve the filtrate for step 4. 

White precipitate indicates strontium. 
Confirm the presence of strontium with a flame test. 
Crimson flame confirms strontium. 

4. Heat the filtrate from step 3 to boiling, and add an 0.5AT solution 
of (NH 4 )2C 2 04 until no further precipitation oceurs. Filter, and 
reserve the solution for the analysis of Group V. 

White precipitate indicates calcium. 

Perform a flame test as confirmation of the presence of calcium in the 
precipitate. 

Brick-red flame proves calcium. 

ANALYTICAL TABLE VIII 
ANALYSIS OF GROUP TV 

The filtrate from Analytical Table V (separation of Group III) may contain 
Ba+ + , Sr ++ , and Ca ++ , as well as the Group V cations. Evaporate, filter and dis- 
card the precipitate. Treat with coned. HCl, and evaporate to a small volume. 
Repeat the treatment, and evaporate to dryness. DO NOT IGNITE. Dilute 
with H 2 0, and again evaporate to dryness. Take up the residue in 1 ml. of Q.2N 
HCl (1). Heat the solution to boiling, and treat with Q.5N (NH 4 )2CrO 4 solution 
(2). Filter. 



Precipitate: BaCr0 4 
(yellow). Confirm 
Ba ++ with flame test 
(2). 


Filtrate: Sr++, Ca ++ . Heat, and treat with 0.5AT 
(NH 4 ) 2 SO 4 solution. Filter (3). 


Precipitate : SrSO 4 (white). 
Confirm Sr++ with flame 
test (3). 


Filtrate: Ca++. Boil, and 
treat with 0.5JV (NH 4 ) 2 - 
C 2 O 4 solution. Filter, 
and reserve filtrate for 
Group V analysis. Con- 
firm Ca ++ in precipitate 
with flame test (4). 



GROUP V CATIONS 

MAGNESIUM, Mg 

Magnesium is a silvery white metal that is so light that it is now 
replacing aluminum in the manufacture of lightweight alloys. Since it 
burns with a brilliant white flame, it is used in flashlight powders in 



208 QUALITATIVE ANALYSIS 

photography. Magnesium falls in Group 2 of the periodic table and is 
therefore an alkaline earth element. 

Magnesium hydroxide, Mg(OH) 2 , is a strong base, although it is 
very insoluble in water. It is, however, soluble in ammonium salts. 
The carbonate, MgC0 3 , behaves similarly in its solubility. Almost all 
magnesium compounds are colorless. There are known, however, 
the so-called lakes, which are highly colored. Notable among these is 
the "lake" obtained with magnesium hydroxide and ^-nitrobenzene- 
azo-resorcinol, which is colored blue. With ammonium and phosphate 
ions, magnesium precipitates as white magnesium ammonium phos- 
phate, MgNH 4 PO 4 , which is quite insoluble in water. 

PRELIMINARY EXPERIMENTS 

1. Cautiously add NH 4 OH to a portion of Mg ++ test solution. To 
another portion of the test solution, add one-half its volume of a 
saturated solution of NH 4 C1, and then add NH 4 OH as before. 

2. To a few drops of the neutral or slightly acid test solution, add a 
drop of p-nitrobenzene-azo-resorcinol solution. Make the solution 
strongly alkaline with NaOH. Run a blank test simultaneously. 

3. Acidify a portion of the test solution with dilute HC1, and heat 
to boiling. Add Na 2 HP04 solution to the hot solution. Add dilute 
NH 4 OH equal to one-third the volume of solution, and allow to cool. 

4. (Optional) Test some "milk of magnesia" for magnesium ion by 
any one of the preceding methods. 

POTASSIUM, K 

Potassium belongs to the family of alkali metals comprising Group 1 
of the periodic table. Freshly cut surfaces of the metal are silvery 
white. Potassium reacts violently with water, with the evolution of 
hydrogen and the formation of the strong, soluble hydroxide, KOH. 
The metal itself is not used commercially to any great extent, but the 
chloride and sulfate are used as fertilizer ingredients. 

Most of the potassium salts are soluble. With perchlorate ion, 
white potassium perchlorate, KC1O 4 , is precipitated. Picric acid yields 
yellow potassium pier ate; a mixture of silver nitrate and sodium 
cobaltinitrite precipitates yellow dipotassium silver hexanitritocobal- 
tate, K2AgCo(NO 2 )6- Bismuth nitrate and sodium thiosulfate yield a 
yellow precipitate of K 3 Bi(S 2 3 )3, tertiary potassium trithiosulfato- 
bismuthite. A yellow precipitate of potassium chloroplatinate, 
K 2 PtCl 6 , is obtained with the chloroplatinate ion. Other insoluble 
salts are also known. 



GROUP V CATIONS 209 

Potassium salts give a characteristic violet flame test which, when 
viewed through a cobalt glass, is red-violet. 

PRELIMINARY EXPERIMENTS 

1. Add perchloric acid, HC10 4 , to about 0.5 ml. of K+ test solution. 
(CAUTION : Do not heat.) 

2. To another portion of the test solution, add a saturated solution 
of picric acid. 

3. To 1 drop of a neutral or slightly acid (acetic acid) solution of 
K + , add a drop of an 0.05% solution of AgNOa and a small amount of 
solid Na 3 Co(NO 2 )6. 

4. Add Carnot's reagent to the test solution. 

Carnot's reagent: Mix 1 drop of 0.5JV Bi(N0 3 )3 with 2 to 3 drops of 
Q.5N Na2S 2 O 3 and 10 to 15 ml. of absolute alcohol. Any turbidity is removed 
by adding a very small amount of H 2 O. 

5. Perform a flame test on the test solution, viewing the flame 
through two or three thicknesses of cobalt glass. 

6. (Optional) Plants contain appreciable amounts of potassium 
salts. Test the ash of a cigarette for potassium by means of the flame 
test. Moisten the ash with concentrated HC1 before touching the 
platinum wire to it. 

SODIUM, Na 

Sodium belongs to the family of alkali metals. The freshly cut 
surface of the metal is silvery white, but it tarnishes very rapidly. It 
reacts rapidly and violently with water, liberating hydrogen and form- 
ing the soluble hydroxide, NaOH, which is a strong base. 

Very few sodium salts are insoluble. With zinc uranyl acetate, 
yellow sodium zinc uranyl acetate, NaZn(UO2)3(C2H 3 O 2 )cr6H 2 O, is 
precipitated. Potassium antimonate precipitates white sodium anti- 
monate, NaH 2 Sb0 4 , formerly erroneously called the pyroantimonate. 

Sodium ion, even in very slight concentrations, gives a yellow flame 
test. One must therefore distinguish carefully between a trace of 
sodium (present as impurity) and an appreciable amount. 

PRELIMINARY EXPERIMENTS 

1. Add some fairly concentrated Na + test solution to the zinc 
uranyl acetate reagent. 

2. Perform a flame test on the test solution. Repeat the test on a 
solution of NH 4 C1. 

3. (Optional) The human body excretes, among other things, 
sodium chloride. Test for the sodium by rubbing a clep-n, cold plati- 
num wire over the palm of the hand, and hold in the flame. 



210 QUALITATIVE ANALYSIS 



AMMONIUM, 

The ammonium ion behaves as if it were the ion of an alkali metal. 
In fact, amalgams of ammonium may be obtained by electrolyzing a 
solution containing ammonium ions, with the use of a mercury cathode. 
Like sodium and potassium, most of the ammonium salts are soluble. 
Picric acid yields yellow ammonium picrate, NH 4 OC 6 H 2 (N02)3. The 
perchlorate ion precipitates white ammonium perchlorate, NH 4 C10 4 . 
Sodium cobaltinitrite produces a yellow precipitate of the correspond- 
ing ammonium salt, (NH 4 ) 3 Co(N0 2 )6, with ammonium ion. Most of 
the common ammonium salts decompose or sublime on heating. 
Ammonium salts are used extensively as fertilizers. 

Ammonia gas, NH 3 , is extremely soluble in water, giving a solution 
that is basic in its reactions and that contains ammonium and hydroxyl 
ions in equilibrium with dissolved ammonia and ammonia vapor. 

NH 4 + + OH- ^ NH 3 + H 2 ^ NH* , 

ivapoi I 
(dissolve,!) 

Since heat usually decreases the solubility of gases in liquids, heating a 
solution of ammonium hydroxide will shift the equilibrium to the right 
and liberate ammonia gas. Strong bases, like potassium and sodium 
hydroxides, will also cause a shifting of the equilibrium to the right, 
displacing ammonia vapor. 

PRELIMINARY EXPERIMENTS 

1. Add dilute NaOH to a portion of NH 4 + test solution, and warm 
gently. Test the vapors with moist litmus paper. 

2. Heat some dry NH 4 C1 strongly in a crucible. Hold a test tube 
filled with water in the vapors. 

3. Treat a portion of the test solution with a saturated solution of 
picric acid. 

4. Add some HC10 4 to another portion of the test solution. 

5. To another portion add some freshly prepared Na 3 Co(N0 2 )6 
solution. 

6. (Optional) Test the white dust that forms on acid bottles in the 
laboratory for the NH 4 +. What is this dust? 

ANALYSIS OF GROUP V CATIONS 

OUTLINE OF SEPARATIONS 

No group reagent is employed for the separation of the Group V 
cations. The ammonium ion is identified in a portion of the original 



GROUP V CATIONS 211 

unknown solution. Magnesium is tested for in one portion of the 
filtrate from Group IV; the other portion of the solution is evaporated 
to dryness and ignited to remove ammonium salts. Tests are then 
made on separate portions of the ammonium-free solution for sodium 
and potassium ions. 

ANALYTICAL PROCEDURE 

1. To a portion of the filtrate from Group I V,' acid with HC1, add 
a drop of p-nitrobenzene-azo-resorcinol, and make strongly alkaline 
with NaOH. 

Blue "lake" indicates magnesium. 

Heat a second portion of the acid solution to boiling, and add Na 2 HP0 4 
solution. Treat the solution with a volume of dilute NH 4 OH equal 
to one-third of the volume of the solution, and allo\\ to cool. 

White crystalline precipitate proves magnesium. 

2. Evaporate the remainder of the Group IV filtrate to dryness, and 
ignite until no more fumes are evolved. Take up the residue in 2 ml. 
of water, and separate into two portions. 

3. Treat a portion of the solution with 1 drop of an 0.05% solution 
of AgNOs, and add some solid Na 3 Co(N0 2 )6. 

Yellow precipitate indicates potassium. 

Perform a flame test on the solution, using several cobalt glasses. 
Rod-violet flame proves potassium. 

4. To the second portion of the solution add some zinc uranyl 
acetate reagent. 

Yellow precipitate indicates sodium. 

Perform a flame test on either the solution or the precipitate. 
Intense yellow flame proves sodium. 

5. Treat a portion of the original unknown with an excess of NaOH, 
and warm gently. Test the vapors with moist litmus paper. 

Odor of NH 3 and red litmus turning blue proves NH 4 + . 



212 



. QUALITATIVE ANALYSIS 



ANALYTICAL TABLE IX 

ANALYSIS OF GROUP V 

The filtrate from Analytical Table VIII may contain Mg++, K + , and Na+, 
and ammonium salts. 



Test a portion of the 
original unknown for 
NEU* with excess 
NaOH (6). 



Make a portion of filtrate from Analytical Table VIII 
acid with HC1. Treat with p-nitrobenzene-azo-resor- 
cinol, and make strongly alkaline with NaOH. Con- 
firm Mg ++ by the phosphate test (1). 



Evaporate remainder of filtrate from Analytical Table 
VIII to dry ness, and ignite. Take up residue in 
and divide into two portions (2). 



Test one portion for K + 
with sodium cobaltini- 
trite. Confirm with 
flame test (3). 



Test second portion for 
Na + with zinc uranyl 
acetate reagent. Con- 
firm with flame test (4). 



ANALYTICAL TABLE X 

GROUP SEPARATION FOR GENERAL UNKNOWN 

Treat 'the not too strongly acid unknown with HC1 until precipitation is com- 
plete. Filter. 



Precipitate : 

Chlorides of 
Group I cat- 
ions. Proceed 
as in Analyt- 
ical Table I. 



Filtrate: Groups II, III, IV, and V. 
saturate with II 2 S. Filter. 



Adjust the acidrty, and 



Precipitate : Group 
II sulfides. Pro- 
ceed as in Ana- 
lytical Tables II, 
III, and IV. 



Filtrate: (H 2 S-free). Groups III, IV, and 
V. Make ammoniacal, add solid NH 4 C1, 
and saturate with H 2 S. Filter, boil fil- 
trate with glacial HC 2 H 3 O 2 . 



Precipitate: Sulfides 
and hydroxides of 
Group III. Pro- 
ceed as in Analyt- 
ical Tables V, VI, 
and VII. 



Filtrate: Groups IV 
and V. Evaporate, 
treat with HC1 sev- 
eral times, and evap- 
orate to dryncss. 
Dilute with H 2 O, 
evaporate to dry- 
ness again, and dis- 
solve residue in 
0.2AT HC1. Pro- 
ceed as in Analyt- 
ical Table VIII, 
using filtrate from 
last step for Group 
V analysis (Ana- 
lytical Table IX). 



CHAPTER XIV 
INTRODUCTION TO BLOWPIPE ANALYSIS 

Blowpipe analysis, which is truly a micro method, has long been 
known and used. In recent years, however, it has fallen into disuse. 
Nevertheless, the method has advantages that make its inclusion in an 
elementary course desirable. Blowpipe analysis is essentially a deduc- 
tive method; i.e., it is not generally feasible to group the ions syste- 
matically or to carry out individual separations, and the student must 
of necessity draw his conclusions from observations made on a limited 
number of tests. Since blowpipe unknowns contain both cations and 
anions, the deductive method is first applied to the rations, with whose 
chemistry the student is already familiar. Subsequently, it is applied 
to the detection of the simpler anions and so serves as an introduction 
to the purely deductive methods applied in the usual schemes of anion 
analysis. 

The method has other marked advantages in a complete analysis of 
both simple and complex materials. First, it often enables the analyst 
to eliminate many specific ions and so shortens considerably the course 
of the analysis. Further, if the unknown is simple, it may be com- 
pletely analyzed by this technique. The method also furnishes many 
excellent specific tests that, indeed, are frequently used even in the 
so-called wet analysis (cobalt bead test, Rinman's Green test for zinc, 
etc.). 

The number of experiments in a preliminary analysis should be few, 
as should the number of reagents. Further, the complete examination 
should take only a short time. The few tests specified carry out this 
principle. It is, of course, possible to verify the presence of many 
substances by other methods, particularly those used in the wet 
method of analysis. However, the student is strongly advised to per- 
form the blowpipe unknowns using only the reagents and experiments 
outlined in the following tables. 

The tests used are: 

1. Heating in a matrass. 

2. Heating (a) on a charcoal block; (6) on a plaster of Paris tablet. 

3. Borax and phosphate beads. 

4. Flame tests. 

[213 



214 



QUALITATIVE ANALYSIS 



5. Treatment with concentrated and dilute acids and alkali. 
The various gases evolved and sublimates formed are usually further 
tested with simple reagents. 

By extending this list, it is possible to do a complete analysis solely 
by blowpipe methods, as was frequently the case in the earlier days of 
qualitative analysis. However, the technique and knowledge of 
chemistry involved are generally beyond the capacity of the beginning 
student. 

HEATING IN THE MATRASS 

The matrass is a tube 3 to 4 in. long and 7 mm. outside diameter, 
having a bulb about 12 mm. in diameter at one end. Although a hard 
glass tube (Pyrex) is preferable, tubes 
made of soda-lime glass suffice. The . 
student should make a number of these 
tubes. The bulb should not have too 



--7mm. 
tubing 



-Test liquid 



\ 



^11 mm. 
FIG. 53. A matrass. 




FIG. 54. Testing for emergent gases. 



heavy walls, and the glass in the bulb should be of equal thickness 
throughout. 

A small amount of material is placed in the bulb and the tube heated 
in the flame. The following observations should be made: 

1. Color of heated substance, hot and cold. 

2. Condensation of vapors in the upper portion of tube. 

3. Sublimates. 

4. Evolution of gases, color, and odor. 



INTRODUCTION TO BLOWPIPE ANALYSIS 



215 



ANALYTICAL TABLE XI 

HEAT IN MATRASS 



Observation 



Decrepitation . 



Carbonization, with burned odor 

and H 2 O 

Water evolved 



Gas evolved 

Colorless and odorless. 



Colorless with characteristic 
odor 



Colored with characteristic odor 
Red-brown 



Reddish. 



Green. 



Violet. 



Sublimate forms 
White 



Yellow hot, white cold. 
Yellow 



Yellow 

Black, with garlic odor. . . 
Black, with violet fumes. 



Gray to gray-black. 



Test 



Action toward litmus: neutral 

Alkaline 
Acid 

Supports combustion Oz 



Burns with a blue flame CO 

Turns baryta water bead tui bid 
COa 

Alkaline to litmus Nils 



Turns baryta water turbid. 
Bleaches KMnO 4 SO 2 

Turns lead acetate paper black 
H 2 S 

Acetone recognized by odor 

HCN recognized by odor 
(CAUTION) 

Starch KI paper turns blue 

-NOz 
Starch KI paper tuiris blue 

Br 2 

Starch KI paper turns blue 
C1 2 

Starch KI paper turns blue 1 2 



(NOTE: may only be color of 
mass) 



Turns red on rubbing 



Indication 



Pb(NOs)2, NaCl and many 
other substances 

Organic matter, as tartrates 
Hydrates, many hydroxides, 

free oxalic acid. 
Ammonium compounds 
Volatile acids 

Seme nitrates, chlorates, perox- 
ides, iodates, certain oxides 

Oxalates 

Carbonates, oxalates, organic 
matter 

Ammonium salts, some organic 
compounds 

S'llfur, sulfides, sulfites, some 
aulfates 

Moist sulfides 



Acetates 
Cyanides 



Nitrates of heavy metals; some 
nitrites 

Bromides in presence of oxidiz- 
ing agents; bromides of some 
metals 

Chlorides of some metals; 
chlorides in presence of oxidiz- 
ing agents 

Iodides of some metals; iodides 
in presence of oxidizing agents 

Ammonium salts, HgiCla, 
HgCla, AszOa, As0 6f SbOi 
ZnO, Zn compounds 

AfiiSi, HgO (also globules of 

Hg), S from sulfides 
Hglz 

Many As compounds 
Iodides in presence of oxidizing 

agents; some iodides 
Oxygen compounds of Hg,HgS 



216 



QUALITATIVE ANALYSIS 



Testing of Emergent Gases. In testing with liquid reagents, a drop 
of the liquid in the loop of a platinum wire, which is held over the 
mouth of the tube, may be used, A better method, however, is to seal 
one end of a small piece of tubing, introduce a drop of the reagent into 
the open end, and hold over the mouth of the tube. 

Testing of the Residue. The residue in the matrass may be 
obtained by breaking the tube. 

NOTE: It is not practical to clean matrasses and reuse them. 

ANALYTICAL TABLE XII 

HEAT IN MATRASS WITH 3 TIMES THE WEIGHT OF ANHYDROUS 
COMMERCIAL Na-2CO 3 



Observation 



Sharp odor 

Gray mirror and globules . 

Gray metal (no globules) 
and garlic odor 



Test 



Alkaline to litmus NH 3 
Rub with filter paper and 
droplets form 



Indication 



All ammonium compounds 
Hg compounds 



As compounds 



REDUCTION ON THE CHARCOAL 

With a charcoal borer or a penknife, make a small, shallow depres- 
sion in a charcoal block. Tamp into the depression a thorough mixture 

of 1 part of substance and 2 parts of 
anhydrous sodium carbonate. Heat in the 
blowpipe flame (using a blowpipe tip on 
the burner), holding the block in an inclined 
position. 



A if In 
Blowpipe w 



Reducing 
flame 




^Oxidizing 
flame 



Blowpipe -tip'* 



Bunsen burner** 




TECHNIQUE NOTES 



1 . Sometimes it is advantageous to place 
a layer of pure sodium carbonate over the 
mixture to keep it from blowing out of the 
cavity; or the mixture may be moistened 
with a drop of water. 

2. With substances reduceable only 
with difficulty, it often helps to mix the 

FIG. M.-The blowpipe flame. fuse with pow dered charcoal. 

3. Heating usually should commence at the far side of the cavity, 
just outside of it, and the block should be gradually moved till the 
substance is in the direct blowpipe flame. 



INTRODUCTION TO BLOWPIPE ANALYSIS 217 

4. To observe sublimates, heat gently, and observe the parts of 
the charcoal that the flame has not touched. The parts of the charcoal 
the flame plays on will often have a gray-to-white deposit of ash on the 
surface, which must not be mistaken for a sublimate. 

5. To test a metallic globule for malleability, clean the pellet, 
place it on the anvil, and strike with the hammer. A malleable par- 
ticle will form a leaflet of metal ; a brittle one will shatter. 

6. In using the blowpipe, it is essential to learn to breathe at the 
same time that one is blowing into the blowpipe. 

ANALYTICAL TABLE XIII 

HEAT SUBSTANCE, WITH BLOWPIPK, ON A CHARCOAL BLOCK 

Observation Indication 

Heated without Na 2 C0 3 

Fuses and runs into charcoal Salts of alkali metals 

Decrepitates NaCl, etc. 

Deflagrates Chlorates and nitrates 

Infusible residue moistened with H 2 is alkaline . . . Ba, Sr, Ca, Mg compounds 
Residue -f- 1 drop of very dilute Co(NO 3 )2 
solution, heated strongly, gives 

Blue mass Al compounds 

Green mass Zn compounds 

Pink mass Mg compounds 

Heated with Na 2 CO 

Metallic globule without incrustation 

Yellow Au salt 

Red Cu salt 

White, malleable Ag salt 

Dissolve in a drop of HNO 3 , add 1 drop of HC1: 
white curdy precipitate 
Metallic globule with incrustation 
Malleable globule 

Soluble in HNO 3 Pb salt 

White ppt. with HN0 3 Sn salt 

Brittle globule 

White incrustation Sb compound 

Yellow to white incrustation Bi compound 

No globule, only incrustation 

Yellow hot, white cold Zn salt 

White with garlic odor As salt 

Colored (see analytical table XIV) 
No incrustation 

Green mass ". Cr compounds 

Black mass (magnetic) Fe, Co, Ni compounds 

White glowing mass Ca, Sr, Mg, Al compounds 

Black mass Sulfates, sulfides, thiosul- 

Placed on silver coin and moistened, gives fates, sulfites 
black stain (Hepar Test) 



218 



QUALITATIVE ANALYSIS 



IODIDE COATINGS ON PLASTER OF PARIS TABLETS 

The plaster of Paris tablet should be prepared beforehand by 
spreading a J^- to }-m. layer of plaster of Paris on an oiled glass sheet 
or oiled Alberene table top. When set, the plaster of Paris layer is 
divided into 3- by 2-in. rectangles and then dried. The dry tablets are 
easily removed. 

A shallow cavity is bored in the tablet, and a mixture of substance 
and "bismuth flux" (1 part KI and 1 part sulfur) is tamped into the 
cavity. The fuse is heated with the blowpipe flame. 

TECHNIQUE NOTES 

1. The degree of heating depends on the substance being examined. 
Some substances must be heated gently in the invisible cone of the 
blowpipe flame. Other substances should be heated very strongly. 

ANALYTICAL TABLE XIV 

HEAT ON PLASTER TABLET WITH BISMUTH FLUX (KI + S) 

Observation Indication 

Greenish yellow with brown both toward center and at distance; 
yellow with drop of (NH^S*; nonvolatile ................... Cd compound 



Chrome yellow; turns black when a drop of (NH^Sx is added. . . Pb compound 
Chocolate brown with underlying red; turns orange-yellow, then 

red when exposed to NH 3 fumes ............................ Bi compound 

Orange-red; disappears when exposed to NH 3 fumes ............ Sb compound 




-- Oxidizing flame 



"educing 1 
(Best -tinged 
with ye/hw) 

Point at which to 
jntrodi/ce sub- 
'stances for 
f let me test 



Dinner cone (Co Id) 



FUSIBLE BEADS ON THE PLATINUM WIRE 

Clean the platinum wire by fusing some 
borax on it and then snapping off the molten 
flux. Repeat this until the bead that forms is 
clear and colorless. Make a clear bead on a 
loop in the platinum wire, and touch it, while 
hot, to a very small amount of the powdered 
substance. Reheat first in the oxidizing 
flame, then in the reducing flame. Observe 
the colors of the bead both hot and cold. 



TECHNIQUE NOTES 

1. After heating in the reducing flame, it is 
often useful to cool the bead in the reducing 
Bunsen atmosphere existing in the cold inner cone of 

the Bunsen flame. 
2. With substances that are oxidized only with difficulty, as Mn0 2 , 
the addition of a little potassium nitrate to the bead will aid the 
oxidation. 



FIG. 



56. The 
flame. 



INTRODUCTION TO BLOWPIPE ANALYSIS 



219 



3. Only minute amounts of the substance should be fused in the 
bead. The intensity of the color varies with the concentration, and 
large amounts of substance often make the bead very dark and opaque. 

ANALYTICAL TABLE XV 
BEAD TESTS ACCORDING TO ELEMENTS* 



Element 




Borax bead 


Microcosmic salt bead 


Oxidizing 


Reducing 


Oxidizing 


Reducing 


Chromium . . . 
Iron 


Hot 
Cold 

Hot 
Cold 

Hot 
Cold 

Hot 
Cold 
Hot 

Cold 

Hot 
Cold 

Sc 

Hot 
Cold 


Yellow 
Yellowish 
green 
Deep yellow 
to orange-red 

Yellow 

Green 
Blue 

Blue 
Blue 
Violet 

Reddish 
brown 
Violet 
Reddish 
violet 
dium carbonate 
flam 
Green 
Yellow-green 
or blue 
Silica si 


Green 
Green 

Bottle green 

Pale bottle 
green 

Colorless to 
green 
Opaque red 
with much 
oxide 
Blue 
Blue 
Opaque gray 

Opaque gray 

Colorless 
Colorless 

j bead green o 
B; colorless in ri 

celeton 
Opaque 


Dirty green 
Fine green 

Deep yellow 
to brownish 
red 
Yellow to al- 
most color- 
less 
Green 

Blue 

Blue 
Blue 
Reddish to 
brownish 
red 
Yellow to red- 
dish brown 
Grayish violet 
Violet 

r blue opaque i 
3ducing flame 
Green 
Yellow-green 
or blue 
Silica s! 
(most eas 


Dirty green 
Fine green 

Red-yellow to 
yellow-green 

Almost color- 
less 

Brownish 
green 
Opaque red 

Blue 
Blue 
Reddish to 
brownish 
red 
Yellow to red- 
dish brown 
Colorless 
Colorless 

n oxidizing 

teleton 
ily seen) 

Opaque 


Copper 


Cobalt. . . 


Nickel .... 


Manganese. . 

Mixtures of 
Fe, Cu, Ni 
Co. 
Silicates 


Ag, Pb, Sb, 
Cd, Bi, Zn, 
Ni 





* Molybdenum, titanium, tungsten, uranium, and vanadium, which give colored beads, are not 
included in this table. 



220 



QUALITATIVE ANALYSIS 



ANALYTICAL TABLE XVI 
BEAD TESTS ACCORDING TO COLORS 



Color 


Borax head 


Microcosmic salt bead 


Oxidizing 


Reducing 


Oxidizing 


Reducing 


Opaque gray 





h.c. Ni, Ag, 




Ag, Pb, Sb, Cd, 






Pb, Sb, Cd, 




Bi, Zn, Ni 






Bi, Zn 






Violet 


h. Ni 




h.c. Mn 






h.c. Mn 








Blue 


c.- Cu 


h.c. Co 


h.c. Co 


h.c. Co 




h.c. Co 




c. Cu 




Green 


c. Cr 


h.c. Cr, Fe 


h.c. Cr 


h.c. Cr 




h. Cu 


h. Cu 


h. Cu 


h. Fe, Cu 


Yellow to 


h. Cr, Ag 




h.c. Fc 


h. Fe 


brownish 


h.c. Fe 




c. Ni 


c. Ni 




c. Ni 




h. Ag 




Red 


h. Fe 


c. Cu 


h.c. Ni 


c. Cu 




c. Ni 


(opaque) 


h. Fe 


h.c. Ni 



NOTE. h. = hot 
c. = cold 
h.c. = hot and cold 



FLAME TESTS 



Clean the platinum wire by heating in the Bunsen flame and by 
running concentrated hydrochloric acid over the wire from a capillary 
dropper and reheating until the wire gives no color in the flame. 
Touch the clean hot wire to some of the substance to be tested, and 
hold in the edge of the flame. Moisten with a drop of concentrated 
hydrochloric acid from the capillary dropper, and reheat. 

TECHNIQUE NOTES 

1. Some substances (notably sulfates) are nonvolatile in the Bunsen 
flame. Such substances should first be heated strongly in the reducing 
flame, then moistened with HCl'and reheated. 

2. In using a cobalt glass in testing for potassium, it is often advisa- 
ble to use three or four thicknesses of glass. 



INTRODUCTION TO BLOWPIPE ANALYSIS 



221 



ANALYTICAL TABLE XVII 

FLAME COLORATIONS 
Compounds Coloration 

Sodium Yellow 

Potassium Violet (through two or three thick- 
nesses of cobalt glass) 
Calcium (volatile compounds, especially 

chlorides) Orange to brick-red 

Strontium Crimson or carmine red 

Barium Yellowish (apple) green 

Antimony Pale greenish white 

Copper halides Blue or green 

Bismuth Pale greenish white 

Lead Pale azure tinged with green 

Zinc Bluish green (usually streaks or 

threads) 

Boron (borates) Green 

Arsenic Pale blue 



TREATMENT WITH ACIDS OR ALKALI 

A small amount of the substance is treated with the reagent in a 
small test tube. The color of any evolved gases is best observed 
against a white background. 

ANALYTICAL TABLE XVIII 
GASES EVOLVED ON TREATMENT WITH HYDROCHLORIC ACID 



Gas evolved 


Tests 


Indication 


C0 2 


No odor; turbidity with baryta water 


Carbonates 


S0 2 


Choking odor; bleaches drop of KMnCh; tur- 


Sulfites 


S0 2 and ppt. of S. . . 
H 2 S 


bidity with baryta water 
Characteristic odor; lead acetate paper turns 


Thiosulfates 
Sulfides 


N0 2 


black 
Brown gas with characteristic odor; starch- 


Nitrites 


HCN 


KI paper turns blue 
CAUTION : bitter almond odor 


Cyanides 









ANALYTICAL TABLE XIX 
GASES EVOLVED WITH NaOH SOLUTION AND HEAT 



NH 3 


Distinctive odor; turns 


Ammonium salts; some 




litmus blue 


organic nitrogen com- 
pounds 



222 



QUALITATIVE ANALYSIS 



ANALYTICAL TABLE XX 

TREATMENT WITH COLD CONCENTRATED H 2 SO4 



Observation 



Tests 



Indication 



Vapor evolved 

Colorless and odorless . 



Colorless with char- 
acteristic odor. 



Characteristic odor and 

color 
Yellowish green . . . 



Yellow 

Brown .to red. 



Violet. 



CO 2 turns baryta water 

turbid 
CO burns with blue flame 

(no blackening) 
Acid vapor turns litmus 

red 
Fumes when breath is 

blown across tube 
Vinegar odor 
Burned odor; blackening 

of solution, SO 2 
Turns water bead turbid; 

etches glass 
Bleaches KMnO 4 bead; 

turns baryta water turbid 



Bleaches litmus; turns 

starch-KI paper blue 

Cl, 

Explosive; sweet odor 
CAUTION: C1O 2 
Chromyl chlorid e 

CrO 2 Cl 2 
Br 2 usually mixed with 

HBr, SO 2 , and H 2 S, 

(test for HBr); starch-KI 

paper 

NO 2 starch-KI paper 
Condenses on upper part 

of tube 1 2 



Carbonates and oxalates 
Oxalates 
Volatile acids 
Halogen acids 

Acetates 

Organic matter, tartrates * 

Fluorides and fluosilicates 

Sulfites, thiosulfates, re- 
ducing agents on H 2 SO4 

Chloride and oxidizing 
agent together 

Chlorates 

Chromates and chlorides 

Bromides 



Nitrates, nitrites 
Iodides 



* NOTE: It is advisable to warm the solution slightly at times, especially in the test for tartrates. 



CHAPTER XV 
ANION ANALYSIS 

GROUP I ANIONS 

CARBONATE, COs" 

Carbonic acid, H 2 C0 3 , does not exist as such but is essentially a 
solution of carbon dioxide in water. Salts of the acid are found widely 
distributed on the earth's surface as limestone, dolomite, marble, chalk, 
and as components of stalactitic and stalagmitic growths in under- 
ground caverns. Soluble carbonates cause water hardness. The gas, 
CO 2 , is widely distributed in the atmosphere, being essential for plant 
photosynthesis . 

The carbonates, which may bo obtained by passing carbon dioxide 
through a solution of a metallic hydroxide, aro gone-rally insoluble, with 
the exception of the alkali metal salts. They undergo hydrolysis and 
form alkaline solutions: 

CO 3 ~ + H 2 O ;= HCO 3 - + OH~ 

A number of carbonates, particularly those of the alkaline earth 
metals, react with excess CO 2 in aqueous solution, forming the soluble 
bicarbonate ion, HCO 3 ~ (primary carbonate ion). The reaction may 
be reversed by boiling the solution, whereupon CO 2 is driven off, and 
the normal (secondary) carbonate is reprecipitated . 

Bicarbonates also hydrolyze to give an alkaline solution: 

HC0 3 - + H 2 ^ H 2 C0 3 + OH- 

but the hydrolysis proceeds to a lesser extent than with the carbonates. 
The alkalinity produced by the hydrolysis of carbonates is detectable 
with phenolphthalein, whereas that of the bicarbonate hydrolysis is not. 

Carbonates decompose with effervescence when treated with 
mineral acids because of the evolution of CO 2 . The gas is colorless 
and odorless and may be recognized by the turbidity produced when it 
is passed into barium hydroxide solution. The disappearance of the 
suspension of BaCO 3 when a large amount of CO 2 is used is due to the 
formation of Ba(HCO 3 ) 2 . 

Barium or calcium chloride precipitates the corresponding white 
carbonate from normal (secondary) carbonate solutions. The 

223 



224 QUALITATIV1 

Stepl 

j 7mm. tubing (soft glass) 

V xa 


V ANALYSIS 

Step? 


'i 


( Of 


Illl 
StepS 


' Illl 

Illl Represents the flame 
Illl (blast lamp y air -gas) 

C Indicates that the tube is rotated 
during particular operation 

A Indicates direction in which force 
I /ff applied 

t Indicates gentle blowing 

Step 2 

When glass is very soft 


VAJ f > ^ ) 


Step 9 




wauea repeciT t ceo 

Step 10 




f Sharp flame 1" 
from bulb -do 
not rotate 

Step 11 


P Vtfry 

, __ ______ \ slowfu 


Final appearance x Seal here and 
/ pull off 


I rr I 


StepS 

fUee to make bulb 




Step 12 
s~\ 


J 


Square end' 
Save for 
side arm 
Step 16 

Step 4 

^ - Pullslowlu 


^ 

Step 13 


co > f 


Step 14 

._^^. 


Final appearance *" 
.'i/ffttf ft'^ffft 


o 


() ^ 

^^-rV_-/ 

f Break off thin bulb a, 
even off inflame 

Step 15 

Final appearance ^-^ 




Step 5 


C ._,_, ,, , , , 
^> 


) 


VnKlsoft Illl 

Step 6 


other view 






Hole should be round and sli$ 



G. 57, Steps in making a micro gas generator. 



Step 16 



GROUP I ANIONS 

Step 22 



225 



Step 17 



Gently 



Step 18 



Joint should be air-tight 
ott this point 

(Proceed rapidly with 
the following steps not 

at/owing the Joint to cool) 



Step 19 B 



Step 20 



Step 21 9 



^ If appearance is not 
smooth and lump of 
glass at Joint has not 
disappeared repeat 
18W19 



Step 23 




Repeat steps 2$ and 21 
untiljoinr is smooth all 
a found the tube. This is 
usual It/ of one in tour steps, 
top, bottom, front & back. 



Step 24- 



Step 25 



Step 26 - 



f 



Anneal by coat ing with 
soot from as moxyf fame 




To make micro-gas 
gene-rater proceed as in 
step 24. To make 502 
generator, bend side arm 
upwards arnd cutoff 
capillary tip. 






? capillary end eft 
is sti/l soft bend 
awards and then 
iff. 




Final appearance 



FIQ. 67. (Continued.) 



226 



QUALITATIVE ANALYSIS 



precipitates are soluble in mineral acids and in excess carbon dioxide. 
Barium carbonate is soluble in dilute (30%) acetic acid, which dis- 
tinguishes it from baritgn sulfite. Silver nitrate precipitates white 
silver carbonate, Ag 2 C0 3 , which turns yellow with an excess of reagent; 
on boiling, the precipitate turns brown because of decomposition into 
Ag 2 O and carbon dioxide. 

PRELIMINARY EXPERIMENTS 

1. Cautiously add dilute HC1 to a small amount of concentrated 
Na 2 C0 3 solution. Prepare a micro gas generator, as illustrated in 
Figs. 57 and 58. Place a small amount of solid Na 2 C0 3 into the bulb, 



: Dropper bulb 



<- Capillary dropper 
Thin rubber tubing 





Micro test tube 
made -from 
7mm tubing 



- Capillary 
delivery 
tip 




Micro Gas 
Generator 



602 Generator 



FIG. 58. Assembled gas generator. 

and fill the dropper with dilute HC1. Add a few drops of acid to the 
salt, and collect the gas in a few drops of Ba(OH) 2 solution. Continue 
until an excess of CO 2 has been added. Boil the resulting solution. 

2. Treat a portion of solid Na 2 C0 3 in the generator with dilute 
HC1. Pass the gas into Ba(OH) 2 solution. Ascertain the solubility 
of the precipitate in 30% acetic acid. 

3. Add a drop of phenolphthalein to a few drops of an 0.01AT 
solution of Na 2 C0 3 . Repeat the test with an 0.01JV NaHCO 3 solution. 
Saturate the Na 2 C0 3 solution with C0 2 , using the microgas generator. 

SULFITE, S0 3 

Sulfurous acid, H 2 SOs, is, in the main, an aqueous solution of sulfur 
dioxide. The solution undergoes auto-oxidation and reduction at room 



GROUP I ANIONS 227 

temperature, forming sulfuric acid and free sulfur. The gas, S0 2 , 
is widely used as a preservative for dried fruits, as a refrigerant, as a 
disinfectant, as an aid in the conversion of wood pulp into paper, and 
in the commercial manufacture of sulfuric acid. 

The alkali metal sulfites are soluble in water; all the others arc 
slightly soluble to insoluble. Since sulfurous acid is dibasic, it also 
forms acid (primary) salts, the bisulfites, most of which arc soluble in 
water. Sulfurous acid and the sulfites are readily oxidized in air or by 
oxidizing agents and are therefore considerably used as reducing 
agents and as bleaches where reducing action is desired. 

Sulfites decompose readily when treated with mineral acids, giving 
off sulfur dioxide, which has a suffocating, pungent odor. The gas 
decolorizes potassium permanganate solution, undergoing oxidation to 
sulfate ion in the process. Sulfurous acid will also decolorize iodine 
solution, reduce arsenate to arsenite, dichromate ion to chromic ion, 
and ferric to ferrous ion. With sodium nitroprusside > Na 2 Fe (CN) 5 NO, 
and zinc sulfate (zinc nitroprusside) and ammonia vapors, sulfur 
dioxide produces a purplish-red coloration. With Ba ++ , sulfites form 
white barium sulfite, BaS0 3 , which is insoluble in 30% acetic acid. 
Strontium ion yields strontium sulfite, which is very slightly soluble in 
water. This fact is used in separating sulfite from thiosulfate. 

PRELIMINARY EXPERIMENTS 

1. Treat a small portion of solid Na 2 SO 3 in a generator with dilute 
HC1. Pass some of the gas into a very dilute solution of KMnO-j, 
acidified with 1 drop of dilute H 2 S0 4 . Repeat the experiment, using a 
dilute iodine solution in place of the KMnC>4. 

2. Place small amounts of solid Na^SOs in each of three test tubes. 
Add 3% H 2 2 to the first, K 2 Cr0 4 solution to the second, and a few 
drops of bromine water to the third tube. After 5 min., treat each 
solution with dilute HC1, and note whether any gas is evolved. 

3. Place some solid Na 2 S0 3 in the bulb of a gas generator that has 
the side arm bent upward (Fig. 58). Fill the dropper with dilute HC1, 
and place a wood splint coated with zinc nitroprusside in the side arm. 
Add a few drops of acid to the solid, and heat on a sand bath for about 
5 min. Remove the splint, and hold it over an open bottle of NHaOH, 
This test is interfered with by sulfide. 

4. Treat a portion of solid Na 2 S0 3 in the generator with dilute HC1. 
Pass the gas into Ba(OH) 2 solution. Ascertain the solubility of the 
precipitate in 30% acetic acid. 

5. Add SrCl 2 solution to a portion of SOa" test solution. 



228 QUALITATIVE ANALYSIS 

THIOSULFATE, 8203" 

Free thiosulfuric acid, H 2 S 2 3 , is unknown because of its instability. 
Except for the salts of the alkali metals, all the thiosulfates are insolu- 
ble or are only sparingly soluble in water. Acidification of an aqueous 
solution of a thiosulfate results in the precipitation of sulfur after a 
short time, together with the evolution of sulfur dioxide. 

Sodium thiosulfate, Na 2 S 2 3 , is the most important salt. This was 
formerly and often still is called sodium hyposulfite. This name is 
incorrect, although the term hypo is still used by photographers. 
It is used extensively as a solvent for unchanged silver halides in the 
" fixing" of photographic film, yielding the complex ion, [Ag 2 (S 2 O3)3]^. 
Alkali thiosulfates dissolve many water-insoluble compounds with the 
formation of complex salts. Lead sulfate, calcium sulfate, the silver 
halides, mercurous chloride, lead iodide, and other substances react in 
this fashion. 

Thiosulfates are fairly strong reducing agents. Potassium per- 
manganate, in acid solution, is reduced to Mn + + by SjjOs^, which, in 
turn, is oxjdized to sulfate ion. The visual reaction is a decoloration of 
the solution. Ferric ion and Cr 2 O7 s=B are also reduced by this anion to 
ferrous ion and chromic ion, respectively. 

Silver ion forms a white precipitate of silver thiosulfate, Ag 2 S 2 3 , 
which, on standing, becomes yellow, then brown and finally black, 
because of the formation of silver sulfide, Ag 2 S : 

Ag 2 S 2 3 + H 2 - Ag 2 S + H 2 S0 4 

When an excess of S 2 O 3 == is present, the complex ion [Ag 2 (S 2 O 3 ) 3 ]^ is 
obtained, which, on dilution and boiling, yields S 2 O 3 = , SO 4 =a , 802, 8, 
and Ag 2 S. Strontium chloride, in faintly alkaline solution that is not 
too concentrated, gives no precipitate with thiosulfate, in contrast 
with sulfite ion. 

When S 2 3 ~ is added to a mixture of sodium azide, NaN 3 , and 
iodine, it functions as a catalytic agent, causing the two latter sub- 
stances to react, and bubbles of nitrogen gas are evolved. Only sulfide 
and thiocyanate interfere with this test. 

PRELIMINARY EXPERIMENTS 

1. Add dilute HC1 to a freshly prepared concentrated solution of 
Na 2 S 2 O 3 . 

2. Mix equal volumes of 8203" test solution and O.IN KMn0 4 solu- 
tion. Acidify with dilute H 2 SO 4 . 

3. To 0.5 ml. of Ag + test solution, add S 2 O 3 ~ solution, drop by drop, 
until the precipitate that first forms redissolves. Heat the solution on 
a water bath for a short time. 



GROUP I ANIONS 229 

4. Seal off one end of a 1-in. length of 5-mm. glass tubing. Place 
some NaN 3 -I 2 reagent in the tube, and clamp it in a vertical position 
with the sealed end at the top. Moisten a platinum wire with dis- 
tilled water, dip into finely powdered Na 2 S 2 3 and introduce into the 
reagent. 

5. Add SrCU solution to a portion of S 2 O3 = test solution. Compare 
with the same test under Sulfite. 

SULFIDE, S 33 

Hydrogen sulfide, H 2 S, is a colorless gas with a characteristic 
" rotten egg" odor. It is partially soluble in water, giving a feebly 
acidic reaction. All the hydrogen sulfide is readily expelled when the 
aqueous solution is heated. The gas, when inhaled, is toxic, and since 
it is used extensively in the laboratory as a precipitant for many of the 
cations, care should always be exercised in its use. 

The sulfides of the alkali metals are readily soluble in water; all the 
others, with a few exceptions, are insoluble. The soluble salts are 
hydrolyzed with the evolution of H 2 S and the production of an alkaline 
solution. The sulfides of the alkaline earth metals, which are only 
slightly soluble, hydrolyze to form the acid sulfide^, which are quite 
soluble in water. The sulfides of A1 4++ , Mg ++ , and Cr ++4 " do not 
exist in aqueous solution. 

Sulfide ion, in both acid and alkaline mediums, is a strong reducing 
agent. Nitric acid is reduced to nitric oxide, NO, and S^ is oxidized to 
free sulfur; KMnO 4 solution is bleached by H 2 S, with the formation of 
Mn +4 " and the separation of free sulfur. 

The characteristic odor of H 2 S is a good indication of its presence. 
A piece of filter paper moistened with Pb(C 2 H 3 O 2 ) 2 solution turns black 
on exposure to H 2 S because of the formation of PbS. The use of 
sodium plumbite, Na 2 Pb0 2 , is believed to give a more delicate test. 
The sulfide ion catalyzes the reaction between sodium azide and iodine 
to evolve nitrogen gas (see Thiosulfate). Alkaline solutions of S~ give 
an intense violet coloration with sodium nitroprusside, Na 2 Fe(CN) 5 NO. 

PRELIMINARY EXPERIMENTS 

1. Treat a small crystal of Na 2 S with dilute HC1. Repeat, holding 
a strip of Pb(C 2 H 3 O 2 ) 2 paper over the mouth of the tube. 

2. Repeat the NaNa-l2 test described under Thiosulfate, substitut- 
ing Na 2 S for the Na 2 S 2 3 . 

3. Place several drops of Na 2 S solution on a spot plate, and make 
slightly alkaline with dilute NaOH. Add a drop of sodium nitro- 
prusside reagent. 



230 QUALITATIVE ANALYSIS 

4. Saturate 5 ml. of H 2 with [28, and add a few drops of the solu- 
tion to each of the following reagents: FeCl 3 , K 2 Cr04, KMn(>4, HN0 3 
(dilute and concentrated). 

5. (Optional) Take 1 ml. of a saturated solution of [28, and deter- 
mine how much it may be diluted and still give a brown coloration with 
Na 2 Pb0 2 solution. [The latter may bo prepared by adding NaOH to 
Pb(N0 3 )2 until the precipitate formed redissolves.] 

CYANIDE, CN~ 

Hydrocyanic acid, HCN, is a clear, colorless liquid having a low 
boiling point (26.5C.)- It has an odor of bitter almonds and is 
extremely poisonous. The acid may be obtained by acidification of 
cyanide salts. It is readily soluble in water and hydrolyzes completely, 
on standing, into ammonium formate, HCOONH 4 . It is an extremely 
weak acid, carbon dioxide liberating HCN from its salts. Hydro- 
cyanic acid is used as a fumigant. It occurs naturally in the glycoside 
(amygdalin) of oil of bitter almonds. 

The acid and salts are extremely poisonous, and care must be 
exercised in their use. The cyanides of the alkali, alkaline earth 
metals, and mercuric mercury are water-soluble; all the others are 
insoluble. A number of these insoluble heavy metal cyanides are 
soluble in an excess of CN~, with the formation of complex ions. This 
tendency is especially characteristic of cyanide ion and is utilized in the 
recovery of gold and in electroplating. The ferro- and ferricyanidcs, 
[Fe(CN) 6 ]^ and [Fe(CN) 6 ]=, are the most common examples of this 
type of complex formation. They are utilized in detecting CN~ by the 
formation of Prussian Blue (ferri-ferrocyanide), Fe 4 [Fe(CN) 6 ]3. 

Cyanides may also be converted to thiocyanate, CNS", by heating 
with ammonium polysulfide. The CNS~ is then identified as the 
blood-red ferrithiocyanate complex, Fe(CNS) + +. This test is extremely 
sensitive and is suitable for detecting CN~ in the atmosphere. When 
a mixture of copper acetate and benzidine acetate is exposed to the 
vapors of HCN, a blue coloration is produced through the oxidation 
of the benzidine. 

PRELIMINARY EXPERIMENTS 

(This section may be omitted at the discretion of the instructor) 

1. Treat a portion of CN~ test solution with dilute HC1. Fan 
some of the vapors toward the nose, and smell cautiously. (CAU- 
TION : NEVER smell HCN directly. Dilute with air by proceeding 
as directed.) 



GROUP I ANIONS 231 

2. Add some FcSC>4 solution, and dilute NaOH to a portion of the 
CN~ test solution. Heat to boiling, acidify with dilute HC1, and add 
several drops of FeCl 3 solution. 

3. To a small volume of CN~ solution add some (NH 4 ) 2 S Z , and 
heat on the water bath until a clear solution is obtained. Make the 
solution faintly acid with dilute HC1, and add a few drops of Fed 3 
solution. 

4. Place 0.5 ml. of the test solution in a porcelain crucible, add 
several drops of dilute H 2 S04, and cover the crucible tightly with a 
piece of filter paper. Moisten the paper with a drop of a previously 
mixed solution of Cu(C2H3()2)2 (2 g./liter) and benzidine acetate. 

NITRITE, NO 2" 

The free acid, HNO 2 , has never been isolated and exists only in 
solution. The latter may be prepared by acidifying a cold solution of 
a nitrite with dilute sulfuric acid. The aqueous solution, pale blue in 
color, decomposes to give nitric oxide, NO, nitric acid, HNOs, and 
water. The first product oxidizes in air to give brown fumes of NO 2 . 

All the nitrites arc readily soluble in water with the exception of the 
silver salt, AgNO->, which is only sparingly soluble. Nitrites function 
readily as both oxidanis and reductants. The anion easily oxidizes 
sujfide to sulfur, ferrous to ferric ion, iodide to iodine, and ammonium 
ion to nitrogen. Many other examples are known. In acid solution, 
N0 2 ~~ reduces all the common oxidizing agents, Mn0 4 ~~, Cr 2 7 = , and 
ClOa". The anion, in turn, is readily oxidized to nitric acid. Nitrites 
form complex ions with a number of cations like cobaltic, f cirrous, and 
cupric ions. The most important of these is the cobaltinitrite (hcxani- 
tritocobaltate) ion, [Co(N0 2 ) 6 ]^. 

When solutions of nitrite ion are treated with sulfuric acid and a 
crystal of FeSO4, a brown ring is produced (see Nitrate). Addition of 
NO 2 ~ to an acid solution of brucine sulfate produces an intense red 
coloration. Solutions of nitrous acid have the property of reacting 
with organic bases (amines), producing compounds that will react or 
1 1 couple" with certain other organic compounds to form colored 
substances and dyes. Treatment of an acetic acid solution of naph- 
thylamine with NaN0 2 solution and sulfanilic acid solution produces a 
deep red coloration. This is a very sensitive test for nitrite ion (Peter 
Griess test). 

PRELIMINARY EXPERIMENTS 

1. Treat a small amount of solid NaNO 2 with dilute HC1 or H 2 S0 4 . 
Repeat the tesi^ holding a strip of starch-KI paper in the escaping 
fumes. 



232 QUALITATIVE ANALYSIS 

2. Place several drops of NaN0 2 solution on a spot plate, and add a 
small crystal of FeSC>4. Allow an equal volume of concentrated 
H2SO4 to flow down the side of the depression. 

3. Dissolve a minute amount of brucine (CAUTION) in a few drops 
of concentrated EUSO-j, and add a drop of N0 2 ~ solution. 

4. Peter Gricss Test. Place several drops of NaNO 2 solution on a 
spot plate, and add an equal volume of dilute acetic acid. Treat the 
solution with a slight excess of a previously mixed solution of naph- 
thylamine and sulfanilic acid. 

ANALYSIS OF GROUP I ANIONS 

1. Group Reagent. Dissolve a portion of the unknown in water 
and treat with dilute HC1. Warm gently. Note the odor (CAU- 
TION), and test the escaping gases with starch-KI paper and with lead 
acetate paper. Nitrite is indicated if the starch-KI paper turns blue, 
and the characteristic black precipitate of PbS and an odor of H 2 S 
indicate sulfide. The deposition of free sulfur and an odor of SO 2 
imply the presence of thiosulfate. A bitter almond odor indicates 
cyanide. 

2. Reducing Acids. Make two ml. of the unknown solution acid 
with dilute H 2 S04. Dilute to 5 ml., and then add 0.5 ml. of dilute 
H 2 S04 in excess. Treat with several drops of 0.1 N KMn0 4 . Bleach- 
ing of the reagent indicates the presence of one or more of the following: 
sulfite, thiosulfate, sulfide, and nitrite. The following tests are 
applied if the tests given above indicate that certain anions may be 
present. 

3. Cyanide. Place about 0.5 ml. of the unknown solution in the 
gas generator, add 40 ing. BaCl 2 , 120 mg. of NaHCO 3 , warm, and col- 
lect the gas evolved in dilute NaOH. Add FeSO 4 solution to the 
NaOH solution, heat to boiling, acidify with dilute HC1, and treat with 
several drops of FcCl 3 solution. 

Blue coloration or precipitate proves cyanide. 

4. Carbonate in the Presence of Sulfite and Thiosulfate. Place 
some of the unknown in a gas generator, treat with dilute HC1, and 
pass the gas into Ba(OH) 2 solution contained in a second gas generator. 
A white precipitate indicates carbonate, sulfite, or thiosulfate, alone or 
in combination. Wash the precipitate, remove the wash water, add a 
few drops of 30% acetic acid, and pass the gas into a tube of Ba(OH) 2 
solution. 

White precipitate proves carbonate. 



GROUP II ANIONS 233 

5. Sulfide. Dissolve a portion of the original unknown in water, 
and treat part of the solution with dilute HCL Place a piece of 
Pb(C 2 H 3 02)2 paper in the escaping fumes. 

Black precipitate of PbS indicates sulfide. 

Place a few drops of tho original solution on a spot plate, and make 
slightly alkaline with dilute NaOH. Treat with a drop of sodium 
nitroprusside reagent. 

Intense violet coloration proves sulfide. 

If S" is present, treat the concentrated unknown solution with solid 
CdC0 3 . Shake, and filter off the excess CdC0 3 and CdS that forms. 
Reserve the filtrate for step 6. 

6. Sulfite. Add some IN SrCl 2 solution to a neutral or faintly 
alkaline solution of the original solution or to the filtrate from step 5, 
if S^ is present. Warm, and allow the solution to stand for about 10 
min. A white precipitate indicates carbonate or sulfite. Filter and 
reserve the filtrate for step 7. Place the precipitate in a gas generator 
with the side arm bent upward, and place a wooden splint coated with 
zinc nitroprusside in the side arm. Treat the precipitate with dilute 
HC1, and heat on a sand bath for about 5 min. Remove the splint, and 
hold it over an open bottle of NEUOH. 

Purple coloration proves sulfite. 

7. Thiosulfate. Treat the filtrate from step 6 with dilute HCL 

Odor of S02 and precipitate of S proves thiosulfate. 

8. Nitrite. If a colored gas was obtained in step 1 and a positive 
reaction was obtained with starch-KI paper, apply the following test: 
Dissolve some of the unknown in water, place a few drops on a spot 
plate, and add an equal volume of acetic acid. Treat with a slight 
excess of a previously mixed solution of naphthylamine and sulfanilic 
acid. 

Deep red coloration proves nitrite. 
GROUP II ANIONS 

FLUORIDE, F- 

Fluorides are found in nature as fluorite, CaF 2 , cryolite, Na 3 AlF 6 , 
and in many silicates. Fluorides in small amounts are essential for 
human metabolism. When larger amounts are ingested, as in certain 
drinking waters, they cause mottling of the teeth. Sodium fluoride, 



234 QUALITATIVE ANALYSIS 

NaF, is used as insecticide and rodent exterminator. Organic fluoro 
compounds (freon) are used as refrigerants in some makes of electric 
refrigerators. 

Although anhydrous hydrofluoric acid, H 2 F 2 , is sold commercially, 
the most common commercial form is the 40% aqueous solution, which 
must be kept in ceresin or wax bottles. The acid is very corrosive to 
the skin and gives very painful burns. Hydrofluoric acid is unique in 
that it will readily react with (etch) glass, silica, and silicates to form a 
gas, silicon tetrafluoride, SiF 4 , .which reacts with water to give silicic 
acid, H 4 SiO 4 , and fluosilicic acid, H 2 SiF 6 . Concentrated sulfuric acid 
liberates the free acid from fluorides. 

Among the metallic salts, silver fluoride, AgF, is soluble in water. 
Barium fluoride, BaF 2 , is soluble in an excess of mineral acid and in 
ammonium salts. Calcium ion precipitates white, slimy calcium 
fluoride, CaF 2 , which is difficultly soluble in hydrochloric acid and 
almost insoluble in acetic acid. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of NaF solution. 
Treat the precipitate with dilute HC1. Treat another portion with 
dilute acetic acid. 

2. Place a small amount of calcium fluoride in a clean lead dish and 
add enough concentrated H 2 SO 4 to make a thin paste. Cover with a 
soft glass slide or glass plate that has been coated with paraffin, 
through which several characteristic symbols have been scratched. 
Let stand overnight. 

3. In a test tube, place a mixture of calcium fluoride and pure 
quartz sand. Add concentrated H 2 SO 4 , and heat. In the mouth of 
the tube, place a glass tube closed at the upper end, containing a drop of 
water at the lower end. 

SULFATE, SO 4 s83 

Sulfuric acid, H 2 S0 4 , is one of the basic substances of industry. 
Although it is made from native sulfur, many sulfates are found in 
nature as gypsum, CaSO 4 -2H 2 O, barite, BaS0 4 , and celestite, SrSO 4 . 
The acid is used in industry whenever a cheap, strong acid is desired. 
The concentrated acid is a very strong dehydrating agent. 

From solutions containing SO 4 = , Ba ++ precipitates white barium 
sulf ate, BaS0 4 , which is insoluble in all acids. Calcium sulfate, CaSO 4 , 
is, however, slightly soluble in water. With concentrated solutions 
of SO 4 -, Ag+ precipitates white silver sulfate, Ag 2 SO 4 , (solubility, 
0.58 g./lOO ml. H 2 at 18C.). With Pb++ white lead sulfate, PbS0 4 , 



GROUP II ANIONS 235 

is precipitated, which is soluble in concentrated ammonium acetate 
solution. 

Sulfates, like all other sulfur compounds, when fused with sodium 
carbonate and charcoal, yield sulfides (Hepar test). Barium sulfate, 
with mercuric nitrate solution, gives a yellow absorption compound. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of Na 2 S0 4 solu- 
tion. Treat the precipitate with dilute HC1. 

2. Hepar Test. Mix thoroughly small amounts of a solid sulfate, 
anhydrous Na 2 CO 3 and powdered charcoal. Place in a cavity in a 
charcoal block and heat with a blowpipe flame. Place the black fused 
mass on a silver coin and moisten with a drop of water. 

3. To a solution of Na 2 SO 4 add excess Pb(C 2 H 3 O 2 )2 solution. 
Treat the precipitate with hot saturated ammonium acetate solution. 

4. To a small portion of barium sulfate on a watch glass add 5 drops 
of mercuric nitrate reagent, and warm gently. 

PHOSPHATE, P0 4 = 

Phosphorus forms several series of acids, the most important being 
the metaphosphoric acid, HP0 3 , pyrophosphoric acid, H 4 P 2 7 , and 
orthophosphoric acid, H 3 PO 4 , in all of which phosphorus has a valence 
number of 5. These acids are derived from P 2 0s by the addition of 
varying amounts of water: 

P 2 5 + H 2 -> 2HPO 3 
P 2 5 + 2H 2 -> H 4 P 2 7 
P 2 O 5 + 3H 2 -> 2II 3 P0 4 

The acid most commonly encountered is orthophosphoric acid, which 
is the only one considered in this analytical scheme. 

Orthophosphoric acid forms three series of salts differing in tho 
number of atoms of hydrogen replaced. For instance, the three 
sodium salts would be primary sodium orthophosphate, NaH 2 PO 4 , 
secondary sodium orthophosphate, Na 2 HP0 4 , and tertiary sodium 
orthophosphate, Na 3 P0 4 . A water solution of the primary salt is acid 
to methyl orange and phenolphthalein. The tertiary salt is basic to 
both indicators; the secondary salt is basic to methyl orange and acid 
to phenolphthalein. 

Orthophosphoric acid is very important, inasmuch as in the form of 
tertiary calcium phosphate it is the main component of the bones of 
vertebrates. Phosphorus is widely distributed in nature in soils and 



236 QUALITATIVE ANALYSIS 

in plants and also occurs in the minerals calcium phosphate and 
apatite. It is a common constituent of fertilizers. 

With Ag + , in neutral solution, yellow silver phosphate, Ag 3 PO 4 , 
is precipitated that is readily soluble in nitric acid and ammonium 
hydroxide. With Ba* 4 ", white amorphous secondary barium phos- 
phate, BaHPO 4 , is precipitated. However, in the presence of NH 4 + , 
tertiary barium phosphate, Ba 3 (P0 4 )2, is obtained. Both barium 
phosphates are soluble in mineral and acetic acids. Ammonium 
molybdate, (NH 4 ) 2 MoO 4 , in excess, slowly precipitates yellow crystal- 
line ammonium phosphomolybdate, (NH 4 ) 3 P0 4 -12Mo0 3 , from nitric 
acid solution. Ferric ion precipitates yellow-white ferric phosphate, 
FePO 4 , insoluble in acetic acid. Magnesia mixture (NH 4 C1, NH 4 OH 
and MgCU) precipitates white crystalline magnesium ammonium 
phosphate, MgNH 4 PO 4 , which is soluble in acids. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of Na 2 HP0 4 solu- 
tion. Treat the precipitate with dilute HC1. 

2. Treat separate portions of primary, secondary, and tertiary 
sodium phosphate solutions with methyl orange and phenolphthalein 
indicators. 

3. Add a large excess of magnesia mixture to an ammoniacal solu- 
tion of a phosphate. ' To the precipitate add dilute HC1. 

4. To a phosphate solution, acid with nitric acid, add a large excess 
of ammonium molybdate reagent, and heat to 50C. 

5. To secondary phosphate test solution, add a slight excess of 
AgNO 3 solution. Treat the precipitate with dilute HN() 3 . 

6. To a solution of sodium phosphate, add FeCl 3 solution. Treat 
the precipitate with acetic acid. 

ARSENATE, As0 4 s 

There are three arsenic acids, corresponding to the three phosphoric 
acids; meta-arsenic acid, HAs0 3 , pyroarsenic acid, H 4 As 2 7 , and 
orthoarsenic acid, H 3 As0 4 . Both the meta and pyro acids react with 
water to give the ortho acid whose salts are, therefore, 'the most 
common. 

Arsenates yield white barium arsenate, Ba 3 (AsO 4 ) 2 , when treated 
with barium salts. The precipitate is soluble in dilute acids. When 
hydrogen sulfide is passed into a cold 0.3JV acid solution of an arsenate, 
no precipitation occurs immediately, although after a long time, 
arsenous sulfide, As 2 S 3 , is precipitated. However, if the solution 
contains a large excess of hydrochloric acid, yellow arsenic penta- 



GROUP II ANIONS 237 

sulfide, As2S5, is precipitated. If the hydrogen sulfide is passed into a 
hot strongly acid solution of an arsenate, a mixture of arsenous and 
arsenic sulfides is obtained. Arsenate is reduced to arsenite by sulfur- 
ous acid. Magnesia mixture (MgCl 2 , NH 4 C1, and NH 4 OH) precipi- 
tates white magnesium ammonium arsenate, MgNH 4 As0 4 , which is 
soluble in acids. A large excess of ammonium molybdate, (NII 4 ) 2 MoO 4 , 
precipitates, from boiling nitric acid solution of arsenate, yellow crystal- 
line ammonium arsenomolybdate, (NH 4 ) 3 As0 4 '12Mo0 3 , which is 
soluble in ammonium hydroxide and alkalis. Iodide ion, in strongly 
acid (HC1) solution, reduces arsenate to arsenite, with the liberation 
of free iodine. Silver ion precipitates chocolate-brown silver arsenate, 
Ag 3 As0 4 (see Arsenic, pages 177 r jf.). 

ARSENITE, AsO 2 ~" 

There are three arsenous acids: meta-arsenous acid, HAs() 2 , pyro- 
arsenous acid, H 4 As 2 Or>, and orthoarsenous acid, HsAs() 3 . The salts 
of the meta acid are those most commonly encountered, although 
orthoarsenite ion exists in solution. 

Barium ion precipitates, from solutions of ar^enite, white barium 
arsenite, Ba(AsO 2 ) 2 , which is soluble in dilute acids. Hydrogen sulfide 
precipitates yellow arsenous sulfide from acid solutions of arsenite. 
Yellow silver orthoarsenite, Ag 3 AsO 3 , is precipitated by the addition of 
silver ion to neutral arsenite solution. Iodine, in neutral solution, is 
decolorized by arsenous acid, with the formation of both iodide and 
arsenate ions (see Arsenic, page 177 'ff.). 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of Na 3 AsO 4 solu- 
tion. Treat the precipitate with dilute HC1. Repeat, using NaAs0 2 
solution. 

2. Saturate a cold solution of NaAsO 2 , slightly acid with dilute 
HC1, with H 2 S. Repeat, using Na 3 As0 4 solution. Treat a hot acid 
(HC1) solution of As0 4 s with solid Na 2 SO 3 , and pass H 2 S into the 
resulting solution. 

3. Make a cold solution of Na 3 As0 4 ammoniacal, and add an excess 
of magnesia mixture. Repeat with a cold solution of NaAsO 2 . 

4. Dissolve a portion of the precipitate from step 3 in dilute HC1. 
Add equal volumes of 0.1N KI solution and CC1 4 and five times this 
volume of concentrated HCL Shake. 

5. Dissolve another portion of the precipitate in dilute HNOs, add 
a large excess of ammonium molybdate reagent, and boil. 



238 QUALITATIVE ANALYSIS 

6. To a solution of Na 8 AsO4, add AgNOs solution. Repeat with 
NaAs0 2 solution. 

TARTRATE, C4H 4 06 I=1 

Tartrates are the salts of the organic acid, tartaric acid, H 2 C4H 4 06. 
The latter occurs both free and as the acid potassium salt in the juice 
of many fruits. It is often found deposited as potassium acid tartrate, 
cream of tartar, KIIC 4 H4()6, in old wine casks. Tartrates, although 
organic substances, arc conventionally included in the inorganic 
schemes of analysis. Potassium acid tartrate is used in cream of 
tartar baking powders. Rochelle salts, sodium potassium tartrate, 
NaKC 4 H 4 06, and tartar emetic, potassium antimonyl tartrate, 
K(SbO)C 4 H 4 6 , are used in medicine. 

Calcium tartrate, CaC 4 n 4 O 67 is soluble in acetic acid and in con- 
centrated alkali hydroxides. Barium ion precipitates white barium 
tartrate, BaC4H 4 6 . On treatment-of tartrates with concentrated sul- 
furic acid, SO 2 , CO 2 , and carbon are formed, the mixture, meanwhile, 
evolving a characteristic "burned-sugar" odor. The decomposition 
may be represented as follows: 



From solutions of neutral tartrates, silver ion precipitates white 
silver tartrate, Ag 2 C 4 H40 6 , which is soluble in nitric acid and ammo- 
nium hydroxide. On warming the ammoniacal solution, black 
metallic silver is precipitated, which, under certain conditions, may 
form a silver mirror on the tube. Tartrates form complex ions with 
many of the metal cations, as aluminum and chromium. They there- 
fore interfere with a systematic analysis and must be removed before 
proceeding to a complete analysis of the metals. , 

When a solution of tartaric acid is treated with potassium chloride 
solution, potassium acid tartrate, KHC 4 H 4 6 , is deposited. This 
reaction is catalyzed by sodium acetate. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl2-CaCl 2 reagent to 1 ml. of NaKC4H 4 6 
solution. Treat the precipitate with dilute HC1. 



GROUP II ANIONS 239 

2. Treat some solid NaKC 4 H 4 6 with concentrated H 2 S0 4 , and 
warm slightly. Test the combustibility of the vapor. Smell the 
vapors cautiously. 

3. Silver Mirror Test. Clean a small test tube thoroughly with 
cleaning solution and distilled water. Add some NaKC 4 H 4 6 solu- 
tion and a slight excess of Agj^Os solution. Add NH 4 OH dropwise 
until the precipitate just dissolves. Add a drop of NaOH, and warm. 

4. To a solution of tartaric acid, add a drop or two of sodium 
acetate solution and some KC1 solution. 

CHROMATE AND BICHROMATE, CrOr*, CrgOy" 

Chromates arc found only rarely in nature. They are usually 
manufactured from chromite ore, FeCr 2 4 , by fusion with sodium 
carbonate in contact with air. Chromates are converted to dichro- 
mates on treatment with acid. 

With Ba++, yellow barium chromate, BaCr0 4 , is precipitated from 
solutions of both Cr0 4 = and Cr 2 7 == . The precipitate is insoluble in acetic 
acid but soluble in hydrochloric and nitric- acids. With Ag + , brick-red 
silver chromate, Ag 2 Cr0 4 , is precipitated that is soluble in HN0 3 and 
NH 4 OH. Lead ion precipitates lead chromate, PbCr() 4 , which is 
soluble in nitric acid and potassium hydroxide. When chromatcs are 
treated with hydrogen peroxide and sulfiiric acid, they arc oxidized to 
blue so-called perchromic acid (CrO 5 ) which is dipcroxy chromium 
oxide, O0(0 2 ) 2 , in which chromium has a valence of 6. This com- 
pound is very soluble in ether. Hydrogen sulfide or alcohol reduces 
chromates to the chromic state: 



16H+ + 2CrOr + 3S~ - 2Cr+++ + 3S + 8H 2 

These reactions may be used to remove chromates. With diphenyl 
carbazide, acid solutions of CrO^ or Cr-jO?" yield a violet color. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of K 2 Cr0 4 solu- 
tion. Treat the precipitate with dilute HC1. 

2. Treat some K 2 O0 4 solution, acidified with dilute HC1, with 
ethyl alcohol, and boil. Pass H 2 S into an acidified (HC1) K 2 CrO4 
solution. 

3. Treat K 2 Cr0 4 solution with AgN0 3 solution. Treat portions of 
the precipitate with dilute HNO 3 and dilute NH 4 OH. 

4. Acidify some K 2 CrO 4 solution with dilute H 2 SO 4 , add 1 ml. of 
ether, then some H 2 2 , and shake well (CAUTION). 



240 QUALITATIVE ANALYSIS 

5. To 2 drops of K 2 CrC>4 solution, acidified with H 2 S04, add an 
equal volume of diphenyl carbazide solution. 

OXALATE, C 2 O 4 ~ 

Oxalic acid is an organic acid that is found in many plants in the 
form of the calcium and acid potassium salts. It is prepared com- 
mercially by heating sawdust and sodium hydroxide. Oxalic acid, 
although organic, is conventionally included in the inorganic scheme of 
analysis. It is used commercially as a cleaning agent for removing 
stains, especially rust spots. Oxalic acid, taken internally, is a violent 
poison. 

Barium oxalate, BaC 2 4 , is soluble in acetic acid. Calcium oxa- 
late, CaC204, is, however, insoluble in acetic acid but is readily soluble 
in hydrochloric or nitric acid. Oxalates will decolorize an acid solution 
of potassium permanganate, giving CO 2 and manganous ion, Mn ++ . 
With concentrated sulfuric acid, oxalates yield equal volumes of C0 2 
and CO. The decomposition may be represented as follows: 



I 



Silver ion precipitates white silver oxalate, Ag 2 C 2 04, which is soluble in 
nitric acid and ammonium hydroxide. When a precipitate of man- 
ganous manganic hydroxide, obtained by the treatment of manganous 
sulf ate with sodium hydroxide, is treated with a solution of an oxalate, 
slightly acid with sulfuric acid, a red coloration, probably due to the 
trioxalato manganite ion, [Mn(C 2 O 4 ) 3 ] s , is obtained. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of (NH 4 )C 2 04 
solution. Treat portions of the precipitate with dilute HC1. 

2. Add a slight excess of AgNO 3 solution to some (NH 4 ) 2 C 2 04 
solution. Treat portions of the precipitate with dilute HNO 3 and 
dilute NH 4 OH. 

3. Treat an acidified (H 2 S0 4 ) solution of (NH 4 ) 2 C 2 4 with dilute 
KMnO4 solution. 

4. Warm gently a solution of a small crystal of manganous sulfate 
in a few drops of water, to which a drop of NaOH has been added. 
Cool, and add dropwise an acidified (H2S04) solution of (NH4) 2 C 2 O4. 
(Interferences: reducing agents.) 



GROUP II AN IONS 241 

BORATE, B0 3 S (B0 2 -, 8407") 

Borates are found in nature chiefly as borax, sodium tetraborate, 
Na 2 B 4 O 7 -10H 2 0. There are several boric acids, orthoboric acid, 
HsBOs, metaboric acid, HB02, and pyroboric acid or tetraboric 
acid, H 2 B 4 7 . The latter is known only in the form of its salts. 
Borax is used as a detergent (cleaning agent), as a flux, and in the 
making of glass. Orthoboric acid is commonly used as an eyewash. 

The salts of orthoboric acid arc unknown in the pure state. The 
borates, therefore, are generally salts of meta- or pyroboric acid. 

Barium ion precipitates white barium metaborate, Ba(B0 2 ) 2 , from 
solutions of borax. This is explained by the fact that a solution of 
borax ionizes as follows: 

Na 2 B 4 O 7 + 3H 2 ^ 2Na+ + 2BO 2 ~ + 2H 3 BO 3 

Upon further dilution, more unionized orthoboric acid is formed, and 
the solution becomes more alkaline. 

B0 2 - + 2H 2 ^ OH - + H 3 BO 3 

Calcium metaborate is soluble in acetic arid. Moderately con- 
centrated solutions of borax give, with Ag+ , a precipitate of white silver 
metaborate, AgB0 2 , which readily changes to brown silver oxide, Ag 2 O, 
on dilution. 

On treatment with concentrated sulfuric acid and alcohols, such as 
ethyl or methyl alcohol, volatile triethyl or trimethyl orthoborate, 
(C 2 H,) 3 BO 3 or (CH 3 ) 3 BO 3 , is formed; this burns with a green flame. 
Boric acid turns turmeric paper reddish brown, which color is stable to 
dilute hydrochloric or sulfuric acid and which turns bluish black if 
moistened with sodium hydroxide solution. Oxidizing agents inter- 
fere with this test. 

PRELIMINARY EXPERIMENTS 

1. Add an excess of BaCl 2 -CaCl 2 reagent to 1 ml. of Na 2 B 4 07 solu- 
tion. Treat the precipitate with dilute HC1. 

2. Treat a concentrated solution of Na 2 B 4 7 with AgN0 3 solution. 
Dilute and warm. 

3. To a small amount of solid Na 2 B 4 O 7 in a crucible, add concen- 
trated H 2 SO 4 and some ethyl alcohol. Ignite the vapors. 

4. Acidify a solution of Na 2 B 4 7 with dilute HCI, and place a drop 
of this solution on turmeric paper. Dry the paper, and then touch 
with a drop of dilute NaOH. (Interferences: chromate, nitrate, 
chlorate, iodide.) 



242 QUALITATIVE ANALYSIS 

SILICATE, Si0 4 ~, SiO^ 

Silicates comprise a large part of the earth's crust. They are 
essential constituents of most rocks and many minerals. Pure silica 
has the formula SiOg. The formulae of the silicates vary, silicon 
assuming a variety of valence numbers. This might be expected since 
silicon falls in the same group in the periodic table as carbon. In the 
various minerals, the silicates are present in the form of giant molecules. 
There are three forms of giant molecules, the thread form, typified 
by asbestos, which has a cross section of one molecule and is x molecules 
in length; the plate form, typified by the micas, which is one molecule 
thick and of undetermined length in the other two dimensions; and the 
massive form, typified by olivine, beryl, etc., which extends endlessly in 
three directions. 

Natural silicates are used as building materials. Many manu- 
factured silicates are also employed in this way, as Portland cement. 
Another important silicate used in industry is silica, SiO 2 , which is used 
in the manufacture of fused silica laboratory apparatus and mercury- 
vapor tubes, and in the manufacture of glass. Glass is a supercooled 
liquid that consists essentially of a mixture of silicates of the alkali 
metals and metals of Group II and 111 of the periodic table. Ordi- 
nary soft glass contains the silicates of sodium and calcium; glass used 
in crystal ware contains large amounts of lead silicate. Hard Bohemian 
glass contains potassium, replacing a large amount of sodium. Pyrex 
glass has as an essential constituent large amounts of boron. Other 
types of glass contain other elements. 

Soluble silicates, like sodium silicate, are used in industry as deter- 
gents (cleaning agents). Sodium silicate is usually sold as a water 
solution. Soluble silicates, when heated with concentrated hydro- 
chloric acid, precipitate silica, SiO 2 . Most silicates, when fused in a 
phosphate bead, leave a silica skeleton in the clear bead. 

ANALYSIS OF GROUP II ANIONS 

1. Group Reagent. Dissolve a portion of solid unknown in water, 
and make alkaline with NH 4 OH. Add some BaCl 2 -CaCl 2 solution, 
stir, and allow to stand for a little while. If the precipitate formed is 
yellow, the presence of chromate is indicated. Treat the precipitate 
with dilute HC1. All the Group II anions are soluble, with the excep- 
tion of BaSO 4 and CaF 2 , the slimy precipitate of the latter being 
slightly soluble. If a precipitate remains, test it for sulfate and fluor- 
ide, as in steps 5 and 8, and test the filtrate for chromate, as in step 4. 



GROUP II ANIONS 243 

2. Oxidizing Acids. Dissolve a portion of the unknown in water, 
and add half its volume of concentrated HC1 and then two-thirds the 
volume of a saturated solution of MnCl 2 in concentrated IIC1. A 
brown or black color indicates the presence of chromate. 

3. Reducing Acids. Acidify a portion of the aqueous solution of 
the unknown with dilute II 2 SO 4 , and treat with a few drops of dilute 
KMnO4 solution. Disappearance of the color indicates the presence of 
one or more of the following anions: AsO 2 ~~, C 2 O 4 ~, or C 4 H 4 O 6 ~. If the 
color does not disappear in the cold, warm gently. 

4. Chromate. If the original precipitate obtained in step 1 is 
yellow, make the filtrate from step 1 ammoniacal, and dissolve the 
precipitate formed in dilute H 2 SOi. Add some H 2 O2 and ether, and 
shake. 

Blue color in ether layer confirms chromate. 

5. Sulfate. Fuse some of the precipitate from step 1 with Na 2 COa 
and powdered charcoal with a blowpipe flame on the charcoal block. 
Place the fused mass on a clean silver coin, and add a drop of water. 

Black stain proves sulfate. 

6. Tartrate. Treat a portion of the solid unknown with concen- 
trated H 2 S0 4 ; warm slightly. 

Blackening of the solid, burned odor, and evolution of H0 2 
indicate tartrate. 

In the presence of reducing agents, it is often necessary to separate 
tartrate before testing. This may be accomplished as follows: Prepare 
a concentrated solution of the unknown, acidify with dilute acetic 
acid, add a drop of NaC 2 H 3 O 2 and some KC 2 H 3 O 2 solution, and let 
stand. A precipitate of KHC 4 H 4 06 is obtained, which is tested for the 
anion by the silver-mirror test. 

7. Oxalate. Dissolve a portion of the original unknown in water 
acidified with acetic acid, and add CaCl 2 solution. The precipitate 
may be either CaF 2 or CaC 2 4 . Filter, and reserve the nitrate for step 
9. Dissolve a portion of the precipitate in dilute H 2 SO 4 , add KMn0 4 
solution, and warm if necessary. 

Bleaching indicates oxalate. 

8. Fluoride. Place the remainder of the precipitate from step 7 
in a clean lead dish, and add sufficient concentrated H 2 SO 4 to make a 
paste. Cover immediately with a soft-glass plate (not a watch glass) 
that has been coated with paraffin, through which a characteristic 



244 QUALITATIVE ANALYSIS 

symbol has been scratched. Allow to stand for a number of hours, 
preferably overnight. 

Etching of the glass proves fluoride. 

NOTE: The use of soft-glass plate like a piece of window glass is recom- 
mended for this test. This type of glass reacts more readily with HF 
than a Pyrex plate or watch glass. 

9. Borate. Evaporate the filtrate from step 7 just to dryness. 
Cool, add concentrated H 2 S04 and ethyl alcohol, and ignite the vapors. 

Green flame proves borate. 

10. Arsenate. Make a cold solution of a portion of the original 
unknown (do not heat) ammoniacal, and treat with a slight excess of 
magnesia mixture. A precipitate indicates the presence of arseriate or 
phosphate or both. Filter, and reserve the filtrate for step 12. Dis- 
solve a small portion of the precipitate in dilute HC1. Add equal 
volumes of 0.1 N KI solution and CC1 4 and five times this volume of 
concentrate^! HC1. Shake. 

Violet coloration in CC1 4 layer indicates arsenate. 

If arsenate is indicated, dissolve the remainder of the precipitate in 
dilute HC1, and add solid Na 2 SO 3 in small portions until all the arsenate 
has been reduced. Saturate the solution with H 2 S. 

Yellow precipitate of As 2 S 3 proves arsenate. 

11. Phosphate. Filter the precipitate from step 10, and boil the 
filtrate to expel excess H 2 S. Cool, add an excess of dilute HN0 3 and 
(NH 4 ) 2 Mo0 4 reagent, and heat. 

Yellow precipitate proves phosphate. 

12. Arsenite. Acidify the filtrate from step 10 with dilute HC1, 
and saturate with H 2 S. 

Immediate precipitation of As 2 S 3 proves arsenite. 
GROUP III ANIONS 

FERROCYANIDE, Fe(CN) 6 a 

The free acid, H 4 Fe(CN) 6 , is a white crystalline solid that is readily 
soluble in water. It is prepared by acidifying a concentrated aqueous 
solution of potassium ferrocyanide, K 4 Fe(CN) 6 . Neither the acid 
nor its salts are found free in nature. The anion is a good example of a 



GROUP III ANIONS 245 

stable complex. Both the acid and its solution are oxidized in air, 
depositing a blue compound that is probably ferri-ferrocyanide. 

Only the ferrocyanides of the alkali and alkaline earth metals aro 
water soluble; all the others are insoluble. The most common salt is 
potassium ferrocyanide. It may be obtained as a by-product in the 
coking of coal. Soluble ferrocyanidos are yellow when hydratod, white 
when in the anhydrous state. Ferrocyanides are mild reductants and 
are readily oxidized in acid solution to ferricyanide, Fe(CN)<r. A 
number of reagents may be used to bring about this reaction: H 2 () 2 , 
Br 2 , Mn0 4 ~, Cr 2 07 =a , HN0 3 , and many others. 

Silver nitrate gives a white precipitate of silver ferrocyanide, 
Ag 4 Fe(CN) 6 , when added to a soluble ferrocyanide. The precipitate 
is insoluble in both dilute nitric acid and ammonium hydroxide. With 
concentrated nitric acid, it is converted to the orange silver ferri- 
cyanide. Ferrocyanides yield, with cobalt nitrate, a green precipitate 
of cobalt ferrocyanide, Co 2 Fe(CN) 6 , which is insoluble in dilute acetic 
and hydrochloric acids. With ferric ion, ferrocyanide ion yields a blue 
precipitate, Fe 4 [Fe(CN) 6 ] 3 , the well-known Prussian Blue. Ferrous 
ion yields a white precipitate that gradually turns blue on exposure to 
air, through oxidation to Prussian Blue. In the presence of acetic 
acid, ferrocyanide ion gives a brown precipitate of cupric ferrocyanide, 
Cu 2 Fe(CN)6, when treated with a solution of CuSO 4 . Lead salts 
produce white lead ferrocyanide, Pb 2 Fe(CN)e, which is insoluble in 
dilute nitric acid. 

Ferrocyanides, when boiled with mercuric oxide, are transposed. 
Prussian Blue, for example, treated with a suspension of yellow mer- 
curic oxide, HgO, is decomposed, yielding ferrous hydroxide, Fe(()H) 2 , 
ferric hydroxide, Fe(OH) 3 , and mercuric cyanide, Hg(CN) 2 . 

PRELIMINARY EXPERIMENTS 

[Use only freshly prepared K^e(CN)^ solution] 

1. Treat 0.5 ml. of K 4 Fe(CN)e solution with a slight excess of 
AgNO 3 solution. Treat the precipitate with concentrated HNO 3 . 

2. Add several drops each of dilute HC2H 3 O 2 and 2N Co(NO 3 ) 2 to a 
portion of Fe(CN) 6 ^ solution. 

3. Acidify a portion of K 4 Fe(CN) 6 solution with dilute HC1, and 
add a few drops of FeCl 3 solution. Treat the blue precipitate with a 
suspension of yellow HgO. 

4. Treat some of the ferrocyanide solution, prepared from a clean 
crystal, with ferrous ammonium sulfate solution. Note the progressive 
color changes of the precipitate on exposure to air. 



246 QUALITATIVE ANALYSIS 

5. Acidify a few drops of K 4 Fe(CN) 6 solution with dilute acetic 
acid, and add some CuS0 4 solution. 

6. Add Pb(N0 3 ) 2 solution to the ferrocyanide solution. Treat the 
precipitate with dilute HNO 3 . 

FERUICYANIDE, Fe(CN) 6 s 

Anhydrous ferricyanic acid, H 3 Fe(CN) 6 , is a brown crystalline 
substance that is readily soluble in water, giving a strongly acid solu- 
tion. The corresponding salts, the ferricyanides, are very stable and 
are generally obtained by oxidation of the corresponding ferrocyanides. 

The ferricyanides of the alkali, alkaline earth metals, and ferric 
ion are water-soluble; all the others are insoluble. Solutions of ferri- 
cyanides are reddish in color, although the ferric salt gives a brownish 
solution. The reaction of ferricyanide ion with various cations pro- 
duces precipitates of varied and pronounced colors. Silver ion forms 
orange-red silver ferricyanide, Ag 3 Fe(CN) 6 , which is insoluble in 
dilute nitric acid but is soluble in ammonium hydroxide. With ferrous 
ion, a deep blue precipitate of TurnbulPs Blue, Fe 3 [Fe(CN) 6 ]2, is 
obtained. When this precipitate is treated with KOH, both ferrous 
and ferric hydroxides are precipitated, and the solution contains ferro- 
cyanide ion. 

A green precipitate of cupric ferricyanide, Cu 3 [Fe(CN) 6 ]2, is 
obtained when Cu ++ is added to soluble ferricyanides. Ferric salts 
produce no precipitate but yield only a brown solution of ferri- 
ferricyanide, Fe[Fe(CN) 6 ]. In contrast to ferrocyanide, no precipitate 
is obtained when ferricyanide solutions are treated with lead salts. 
In the presence of acetic acid, ferricyanide yields,, with Co 4 " 1 ", a reddish 
precipitate of cobalt ferricyanide, Co 3 [Fe(CN) 6 ]2. In acid solution, 
ferricyanide is reduced to ferrocyanide by potassium iodide, KI, which 
is in turn oxidized to free iodine. This reaction forms the basis for the 
quantitative estimation of the anion. As an oxidizing acid, also, 
ferricyanide converts a hydrochloric acid solution of MnCU to the dark 
brown Mn' HH " + ion. 

PRELIMINARY EXPERIMENTS 

[Use only freshly prepared KJ?e(CN)s solution] 

1. Add a slight excess of AgNO 3 solution to K3Fe(CN) 6 solution. 
Treat separate portions of the precipitate with dilute HNO 3 and dilute 
NH 4 OH. 

2. Add a portion of ferricyanide solution to a saturated solution of 
MnCl 2 in concentrated HC1. 



GttOUP III ANIONti 247 

3. Acidify a portion of Fe(CN) 6 ~ with dilute HC1, and add a few 
drops of a freshly prepared solution of ferrous ammonium sulfate. 
Boil some of the precipitate with KOH solution. Filter and acidify the 
filtrate and treat with FeCl 3 solution. 

4. Add a few drops of CuS0 4 solution to an acidified (HC 2 H 3 O2) 
solution of potassium fcrricyanide. 

5. Treat ferricyanide solution with Pb(N0 3 )2 solution. 

6. To an acidified (HC1) solution of ferricyanide, add an equal 
volume of KI solution. 

7. Add several drops of dilute acetic acid and 2N Co(NO 3 ) 2 solution 
to the ferricyanide solution. 

THIOCYANATE, CNS~ 

Thiocyanic acid, HCNS, is an oily liquid readily soluble in water, 
giving an aqueous solution whose strength is about equal to that of 
hydrochloric acid. It is not very stable, however, and decomposes 
with the liberation of HCN. The salts are much more stable. Alkali 
thiocyanates may readily be prepared by heating the corresponding 
cyanide with ammonium polysulfide (see Cyanide*, page 230). 

Practically all the thiocyanates arc* water-soluble, the only excep- 
tions being the silver, mercury, and copper salts. However, even 
these are soluble in excess of CNS~ with the formation of complex ions. 
Silver ion reacts with CNS~~ to form a white, curdy precipitate of silver 
thiocyanate, AgCNS, which is soluble in ammonium hydroxide but is 
insoluble in dilute nitric, acid. The precipitate may be transposed by 
boiling with 5% NaCl solution, yielding silver chloride and a solution 
of sodium thiocyanate, NaCNS. This fact is utilized in the separation 
of AgCNS from Agl. Ignition of silver thiocyanate results in decom- 
position with the formation of black silver sulfide, which reaction may 
be used to separate and distinguish the anion from the halides. With 
ferric ion, a red complex, Fe(CNS) + +, is obtained that is readily soluble 
in ether. Mercuric nitrate, with CNS~, yields a white precipitate of 
mercuric thiocyanate, Hg(CNS) 2 , which is soluble in an excess of KCNS 
with the formation of the complex ion, [Hg(CNS) 3 ]~". The potassium 
salt of the latter, when heated, expands into peculiar shapes (Pharaoh's 
serpents). 

Cupric ion produces an emerald-green solution with thiocyanate 
ions; with an excess of the reagent a black precipitate of cupric thio- 
cyanate, Cu(CNS) 2 , is produced. However, in the presence of sul- 
furous acid, a white precipitate of cuprous thiocyanate, Cu 2 (CNS) 2 , 
is obtained. Thiocyanate ion catalyzes the reaction between sodium 



248 QUALITATIVE ANALYSIS 

azide and iodine with the evolution of nitrogen gas. Sulfide and 
thiosulfate ions also catalyze this reaction. 

PRELIMINARY EXPERIMENTS 

1. Add a slight excess of AgX0 3 solution to about 0.5 ml. of 
NH4CNS solution. Boil a portion of the precipitate with 5% NaCl 
solution for a few minutes. Filter, and treat the filtrate with FeCls 
solution. Treat a portion of the precipitate with dilute NH 4 OH. 

2. Treat NH 4 CNS solution with dilute FeCl 3 . Shake a portion of 
the solution with some ether (CAUTION). To the remainder add 
Hg(N() 3 ) 2 solution. 

3. To some of the CNS~~ solution add lead nitrate solution. 

L Add NH 4 CNS solution dropwise to a little CuSO 4 . Divide the 
solution into two portions. Add an excess of CNS~ to one portion. 
To the other, add a solution of HoSOs, prepared by acidifying a solution 
of Na 2 S0 3 . 

5. Mix 1 drop of NH 4 CNS solution with an equal volume of NaN 3 -Iu 
reagent on a spot plate. 

CHLORIDE, Cl~ 

Hydrochloric acid is an aqueous solution of the extremely 
soluble gas, hydrogen chloride. The concentrated solution contains 
from 36 to 37% HC1. Dilute HC1 is t a very strong acid, being highly 
ionized. The salts of the acid are very common and abundant, par- 
ticularly the sodium and potassium salts. 

Strong oxidizing agents, like Pb() 2 , KMnO 4 , etc., convert the acid 
into free chlorine. Concentrated nitric acid liberates some nitrosyl 
chloride, NOC1, as well as chlorine. When these acids are mixed in the 
ratio of 1 part HNO 3 :3 parts HC1 by volume, the resulting solution is 
called aqua regia and serves as an excellent solvent for gold and the 
platinum metals. 

Most of the metallic chlorides are soluble in water, the exceptions 
being the salts of silver, lead, mercurous mercury, cuprous copper, and 
the basic chlorides of antimony and bismuth. The sodium salt, 
NaCl, is common table salt. A white, curdy precipitate of silver 
chloride, AgCl, is obtained when silver nitrate is added to a soluble 
chloride. The precipitate is soluble in ammonium hydroxide, sodium 
thiosulfate, and potassium cyanide (see reactions of Silver. Cation 
Group I). Silver chloride may be transposed by reduction with metal- 
lic zinc and dilute sulfuric acid, metallic silver being precipitated and 
the chloride ion passing into solution. 



GROUP III ANIONS 249 

Anhydrous chlorides, when treated with potassium dichromate, 
K 2 Cr 2 07, and concentrated sulfuric acid, are converted to the brown- 
ish red gas chromyl chloride, CrO 2 Cl 2 . When the gas is absorbed in 
sodium hydroxide solution, sodium chromate is formed, which is 
tested for in the usual manner. This test confirms the presence of 
chlorides. 

PRELIMINARY EXPERIMENTS 

1. To 1 ml. of NaCl solution add a slight excess of AgNO 3 solution. 
Treat a portion of the white precipitate with an excess of dilute 
NH4OH. Acidify the resulting solution with dilute HNO 3 . Boil the 
remainder of the precipitate with dilute H2S04 and metallic Zn. 
Filter, and test the filtrate for Cl~. 

2. Add a few drops of concentrated H 2 S04 to a few crystals of solid 
NaCl. Hold a glass rod, wet with dilute NEUOH, over the mouth of 
the tube. 

3. Mix a small amount of solid NaCl with twice its weight of 
finely powdered iv 2 Cr 2 O 7 , and place the mixture in a micro gas gen- 
erator. (The apparatus must be perfectly dry.) Arid several drops of 
concentrated H 2 SO4, and heat on a sand bath ior several minutes. 
Collect the gas evolved in dilute NaOH. Make the solution acid with 
dilute H 2 S0 4 , add several drops each of other and H 2 O 2 , and shake. 

BHOMIDK, Br~ 

Hydrobromic acid is formed by solution of the gas, hydrogen 
bromide, in water. It is a strong acid but is not so stable as hydro- 
chloric acid. On exposure to light, it is oxidized with the evolution of 
free bromine, which imparts a yellow color to the solution. In general, 
hydrobromic acid is more easily oxidized than hydrochloric acid. 

The metallic bromides resemble the chlorides, both in solubility and 
chemical behavior. Sodium and potassium bromides are used medici- 
nally as sedatives. Only the silver, mercurous, cuprous, and lead 
bromides are insoluble in water. Silver nitrate precipitates, from 
soluble bromides, curdy yellow-white silver bromide, AgBr, insoluble 
in nitric acid, .less soluble in ammonium hydroxide than AgCl, but 
readily soluble in potassium cyanide and sodium thiosulfate, forming 
complexes of the same type as silver chloride. The precipitate may be 
reduced with zinc and sulfuric acid, or it may be transposed by boiling 
with 5% NaCl solution, yielding, in either case, a solution of the 
bromide ion. Conversion takes place, with NaCl, to the extent of 
about 20%. 



250 QUALITATIVE ANALYSIS 

Oxidation of bromides and hydrobromie acid with oxidizing agents, 
like KMnC>4, results in the evolution of free bromine that is soluble in 
and imparts a brown to red color to carbon tetrachloride, CC1 4 . A 
sensitive test for bromides involves the action of free bromine on the 
organic dye fluorescein. In this reaction, the organic compound is 
converted to the red dye, eosin. The latter is the familiar coloring 
matter of red ink and indelible lipsticks. 

PRELIMINARY EXPERIMENTS 

1. Add a slight excess of AgNO 3 solution to KBr solution. Treat a 
portion of the precipitate with dilute NH 4 OH. Boil another portion of 
the precipitate with dilute H 2 SO 4 and granular Zn, filter, and test the 
filtrate for Br~. Boil the remainder of the AgBr with 5% NaCl solu- 
tion for several minutes. Filter, and test the filtrate for Br~. 

2. Treat some of the KBr solution with several drops of an acidified 
(H 2 S0 4 ) solution of KMn0 4 . Add some CC1 4 to the solution, and 
shake the tube. 

3. Add several drops of concentrated HoSO 4 to some solid KBr. 
Hold a strif) of paper impregnated with fluorescein in the mouth of the 
tube. 

IODIDE, I~ 

Hydrogen iodide, HI, is a colorless gas, with an irritating odor, 
which is extremely soluble in water, forming hydriodic acid. It is a 
strong acid and is highly ionized but is less stable than cither hydro- 
bromie or hydrochloric acid. It is oxidized in air, liberating free iodine, 
which imparts a brown color to the solution. Hydriodic acid is easily 
oxidized by chemical agents even in the cold. 

The metallic iodides of lead, cuprous and mercury ions are more 
insoluble in water than the corresponding chlorides and bromides. 
Sodium and potassium iodides arc used medicinally in the treatment of 
goiter. This disease is caused by lack of the hormone thyroxin, 
an organic iodo compound found in the thyroid gland. 

Silver nitrate forms a yellow precipitate of silver iodide, Agl, which 
is insoluble in nitric acid and is only very slightly soluble in ammonium 
hydroxide. It is, however, readily soluble in potassium cyanide and 
sodium thiosulfatc solutions. Silver iodide may be reduced with zinc 
and sulfuric acid, but it is not transposed by boiling NaCl solution. 
Lead salts precipitate yellow lead iodide, Pbl2, which is sparingly 
soluble, even in hot water. Hydrogen peroxide oxidizes iodide 
solutions to free iodine, which imparts a violet color to carbon tetra- 
chloride, in which it is soluble. Other oxidizing agents bring about the 



GROUP III ANIONS 251 

same reaction. Concentrated sulfuric acid, acting on iodides, evolves 
free iodine. Nitrous acid also liberates free iodine, which turns starch 
solution a blue color. 

PRELIMINARY EXPERIMENTS 

1. Add a slight excess of AgNO 3 solution to KI solution. Test the 
solubility of the precipitate in NH-tOH. Boil the remainder of the 
precipitate with dilute H 2 SO 4 and Zn, filter, and test the filtrate for I~. 

2. Treat a portion of the I~ solution, to which some CC1 4 has been 
added, with several drops of H 2 O 2 . Shake the tube and contents. 

3. Add concentrated HuSC^ to a few crystals of KI, and warm 
gently. 

4. To a few drops of Pb(N0 3 )2 solution, add KI until precipitation 
is complete. (Interference: ferrocyanide.) 

5. Place a drop of acidified I~ solution (HC1) on a spot plate, and 
add equal volumes of starch solution and KNO 2 solution. 

ANALYSIS OF GROUP III ANIONS 

1. Group Reagent. Dissolve a portion of the unknown in water, 
and make the solution distinctly acid with dilute HN() 3 . Add AgN0 3 
solution dropwise until no further precipitation occurs. The presence 
of certain anions is indicated by the color of the precipitate. Ferro- 
cyanide, thiocyanate, and chloride give white precipitates; bromido is 
yellow- white; iodide, yellow; and ferricyanide, orange. 

NOTE: In the event that only Cl~, Br~, and I~ are present, it should be 
noted that Agl precipitates first, then AgBr, and finally AgCl. 

2. Oxidizing Acids. Treat a portion of the unknown solution with 
half its volume of concentrated HC1, and then add two-thirds the 
volume of a saturated solution of MnCl 2 in concentrated HC1. The 
appearance of a brown to black coloration indicates the presence of 
ferricyanide. 

3. Reducing Acids. Dissolve a portion of the unknown in water, 
and acidify the solution with dilute H 2 SO 4 . Treat the solution with a 
few drops of O.1AT KMn(>4 solution. Bleaching of the solution indi- 
cates the presence of one or more of the following anions: Fe(CN) 6 ~, 
Br~, and I~. 

4. Action of Concentrated H 2 SO 4 . Treat a small portion of the 
well-mixed solid unknown with a few drops of concentrated H2SO4, and 
warm gently. The evolution of brown fumes indicates Br~; violet 
fumes indicate I~. 



252 QUALITATIVE ANALYSIS 

5. Test for Ferrocyanide, Ferricyanide, and Thiocyanate in the 
Presence of Each Other. On a piece of good filter paper, place a drop 
of Pb(NO3)2 solution, and allow it to absorb completely. Add a drop 
of the unknown solution to the center of the wet spot. When this 
has absorbed completely, add another drop of Pb(NO 3 )s solution. 
After complete absorption of this last drop, spread the spot by adding, 
singly, 2 to 3 drops of water. The spot should now be about 7 cm. 
in diameter. Ferrocyanide is now precipitated at the center as the lead 
salt. Ferricyanide and thiocyanate have diffused to the periphery of 
the spot. With a capillary dropper, draw a thin line of a solution of 
ferrous ammonium sulfate across the spot. 

A deep blue where the line crosses the periphery of the spot 
proves ferricyanide. 

With another capillary dropper, draw a line across the spot, at right 
angles to the first line, with FeCl 3 solution. 

A red coloration, at the periphery, proves thiocyanate. 
A blue spot in the center proves ferrocyanide. 

NOTE: There may appear occasionally a light blue coloration at the 
center, where the ferrous sulfate line crosses it. This is caused either 
by slight contamination of the ferrous salt with ferric ion, or by air oxida- 
tion of the white ferro-ferrocyanide. For this reason, the ferrous salt 
solution should always be freshly prepared from a clean green crystal of 
ferrous ammonium sulfate. 

6. Cobalt Nitrate Separation. If fcrro- and ferricyanides are 
absent, proceed directly as in step 7. If these anions are present, 
proceed as follows: Acidify a portion of the unknown solution with 
dilute acetic acid. Add an excess of 2AT Co(N0 3 )2 to the solution, 
with vigorous stirring. Treat with approximately 0.1 g. of washed 
asbestos, boil the solution for about 12 min., and filter. Wash the 
precipitate twice with hot water, adding the washings to the filtrate, 
and discard the precipitate. 

7. Make the filtrate from step 6 slightly alkaline with dilute NaOH, 
and then make it distinctly acid with concentrated HNOs. (If ferro- 
and ferricyanides are absent, the original unknown may be acidified 
directly.) Add a slight excess of AgNOs solution, with vigorous stir- 
ring. Filter and wash the precipitate, and discard the filtrate and 
washings. 

8. Bromide. Treat a portion of the precipitate from step 7 with 
5% NaCl solution, and boil for about 5 min. Filter, and discard the 



GROUP IV ANIONS 253 

precipitate. The filtrate contains bromide, or thiocyanate, and chlo- 
ride. Add some CC1 4 and H 2 2 to the solution, and shake. 

Red-brown color in CCU (lower layer) proves bromide. 

9. Separation of Chloride and Iodide. If CNS~ is present, ignite 
the remainder of the precipitate from step 7 until it turns black and the 
burning of sulfur ceases, and proceed as below. If CNS~ is absent, 
proceed directly with the following step. Treat the residue with 
dilute H 2 SC>4 and granular Zn, and boil for 3 to 5 min. Filter, wash the 
precipitate with hot water, and discard the precipitate. 

10. Iodide. Treat a portion of the filtrate from step 9 with H 2 C>2 
and CCU, and shake for several seconds. 

Violet coloration in CCU layer proves iodide. 

11. Chloride. If iodide and bromide aro present, they are removed 
by boiling the remainder of the solution from step 9 with HNOs until 
no more fumes of Br 2 or I 2 are evolved. Treat the solution with a 
slight excess of AgN0 3 

White precipitate proves chloride. 

GROUP IV ANIONS 

NITRATE, N() 3 ~ 

Nitrates are found in quantity in nature, as sodium nitrate (Chile 
saltpeter), in the rainless deserts of Chile and in other parts of the 
world. They are widely distributed in 1he soil as ammonium nitrate, 
which is essential to the growth of plants. Nitric acid, HNOa, is 
manufactured by the oxidation of ammonia formed by combination of 
nitrogen and hydrogen under high pressure and medium temperature 
(Habcr process). Nitrates are used in fertilizers, gunpowder, and in 
many industries. 

Nitrates, in general, are soluble in water, although a few react with 
water to give insoluble oxynitrates (hydroxynitrates). Nitric acid is a 
strong oxidizing agent. With metals, it does not yield hydrogen (with 
some exceptions) but gives, instead, water and some of the lower oxides 
of nitrogen or ammonia, depending upon the conditions and the metal 
used. Concentrated nitric acid renders many metals passive (incap- 
able of reacting). A notable example of this phenomenon is aluminum, 
which is rendered so passive that nitric acid is now shipped in drums 
made of this metal. The concentrated acid forms aqua regia with con- 
centrated hydrochloric acid. This is a mixture of nitrosyl chloride, 



254 QUALITATIVE ANALYSIS 

NOC1, and free chlorine. Concentrated sulfuric acid reacts with 
nitric acid with the liberation of nitrous fumes; dilute sulfuric acid does 
not bring about this reaction, in contradistinction to nitrous acid 



Ferrous ion, acted upon by nitric acid, is oxidized to ferric ion, with 
the evolution of nitric oxide, NO. The oxide combines with more of 
the ferrous ion to form a brown compound, FeSCVNO. This is the 
basis of the so-called brown-ring test for nitrates. Aluminum, 
or zinc, and sodium hydroxide will, on heating, reduce nitrates to 
ammonium ion. Brucine sulfate, in concentrated sulfuric acid, added 
to a solution of nitrate in concentrated sulfuric acid gives a deep red 
coloration which fades through a range of colors to give, finally, a 
greenish yellow color. 

PRELIMINARY EXPERIMENTS 

1. Treat a portion of NaN0 3 with concentrated H2S04. Repeat 
the test with dilute H 2 S0 4 . 

2. Add NaOH solution to a solution of NO 3 ~, and boil. Smell the 
vapors, and hold a piece of moist red litmus paper over the mouth of 
the tube. Cool the tube, add some aluminum dust, and boil again. 
Smell the vapors, and hold a piece of moist red litmus paper over the 
mouth. 

3. Place a tiny crystal of ferrous sulfate or of ferrous ammonium 
sulfate in a depression on the spot plate. Cover with a drop of NaN0 3 
solution, and add a drop of concentrated H 2 S04 down the side of the 
depression. 

4. Mix some NaNO 3 solution with three times its volume of con- 
centrated H 2 SC>4, and treat with a few drops of a solution of brucine in 
concentrated H 2 S04. 

CHLORATE, C10 3 ~ 

Free chloric acid, HC1O 3 , is quite unstable, decomposing into per- 
chloric acid, HC10 4 , and liberating chlorine and oxygen. It is a very 
strong oxidizing agent. The most common salt is the potassium salt, 
KClOs. This compound is used in the manufacture of matches, fire- 
works, and some explosives. 

Although all chlorates are soluble in water, some are hydrolyzed. 
Potassium chlorate is the least soluble of the stable chlorates. Dilute 
sulfuric acid, when added to a chlorate, liberates free chloric acid, 
which decomposes, yielding chlorine, oxygen, and perchloric acid. 
This mixture, will turn starch-KI paper blue. Concentrated sulfuric 
acid reacts with chlorates to form perchloric acid and yellowish green 



GROUP IV ANIONS 255 

chlorine dioxide, C10 2 , which is violently explosive, especially on heat- 
ing. Chlorine dioxide has a peculiar sweetish odor that is quite 
distinctive. 

Chlorates are reduced by aluminum, or zinc powder and sodium 
hydroxide to chlorides. Chlorates may also be reduced in acid solu- 
tion. If a drop of an aqueous aniline sulfate solution is added to some 
sulfuric acid in which a crystal of a chlorate has been placed, a deep 
blue color is obtained. This test is not given by nitrate. 

PRELIMINARY EXPERIMENTS 

1. To a drop of dilute H 2 SO4, on a spot plate, add a tiny crystal of 
KC10 3 . Immerse a strip of starch-KI paper in the liquid. Repeat the 
test, using concentrated H 2 S0 4 . 

2. Make a solution of KC1O 3 strongly alkaline with NaOH, add 
some Al dust, and boil. Filter off the Al dust, acidify the solution with 
dilute HNOs, and add an excess of AgNO 3 solution. 

3. (Optional) Add a small crystal of KC1O 3 to several drops of 
concentrated H 2 S0 4 on a spot plate, and add a few drops of an aqueous 
solution of aniline sulfate. 

ACETATE, C 2 H 3 2 ~" 

Acetic acid, HC2H 3 O2, an organic acid, occurs naturally in many 
plants, both free and as the potassium or calcium salt. It is obtained 
commercially by the dry distillation of wood or by the oxidation of 
ethyl alcohol. Vinegar is a dilute solution of acetic acid obtained by 
the oxidation of the alcohol in various fermented liquors, like wine or 
cider. Acetic acid itself has a very choking odor reminiscent of 
vinegar. The sodium and lead salts are the most important commer- 
cially used acetates. 

Most of the acetates are soluble in water, although the silver salt is 
only sparingly soluble. Dilute sulfuric acid liberates acetic acid from 
its salts, and since the acid is quite volatile, its odor may be readily 
recognized, particularly if the solution is warmed. If ethyl alcohol is 
added to acetic acid or an acetate, concentrated sulfuric acid added, 
and the solution warmed, ethyl acetate, which has a pleasant fruity 
odor, is evolved. Ferric chloride, added to neutral acetate solutions 
in the cold, forms dark brown ferric acetate, Fe(C 2 H 3 2 )3, which, on 
dilution and boiling, precipitates the dark red to brown basic ferric 
acetate, Fe(OH) 2 (C 2 H 3 O 2 ). Acetate ion, with iodine, ammonium 
hydroxide and lanthanum nitrate, yields a dark blue precipitate that 
is probably due to the formation of a complex adsorption compound. 



256 QUALITATIVE ANALYSIS 

When acetates are heated strongly, acetone, (CH 3 ) 2 CO, is evolved. 
When the alkali metal salts are used, the carbonate is left behind; with 
the alkaline earth salts, the oxide remains. 

PRELIMINARY EXPERIMENTS 

1. Add some dilute H 2 SO 4 to an acetate, warm, and smell the 
vapors. 

X 2. To several drops of an acetate solution, acidified with concen- 
trate H 2 SO 4 , add several drops of ethyl alcohol. Warm the solution, 
and smell the vapors. Run a blank in conjunction with this test. 

3. To a neutral solution of NjiC 2 H 3 O 2 , add a small volume of FeCl 3 
solution. Dihitc, and boil the solution. 

4. To a drop of an acetate solution, on a spot plate, add 1 drop of 
5% lanthanum nitrate solution, a drop of an 0.0 IN iodine solution, and 
several drops of dilute NH 4 OH. [Ba(NO 3 ) 2 solution should be added 
to remove sulfate and phosphate, if they are present.] 

ANALYSIS OF GROUP IV ANIONS 

* 

(Only Group IV anions present) 

1. Treat a small portion of the solid unknown with some concen- 
trated H 2 S04. Smell the vapors cautiously. If any of the members 
of this group are present alone, their presence may be indicated by this 
test. 

2. Nitrate. Dissolve some of the solid unknown in water. Make 
the solution alkaline with dilute NaOH solution, and boil to expel any 
NH 3 that may be present. Cool the solution, add some aluminum 
dust, and boil. Place a piece of moist red litmus paper over the mouth 
of the tube. 

Odor of NH 3 , with red litmus turning blue, proves nitrates* 

3. Chlorate. Filter off the aluminum dust from the solution from 
step 2, divide the filtrate into two parts, and reserve one for step 4. 
Acidify the other portion with dilute HN0 3 , and add a slight excess of 
AgN0 3 solution. 

White precipitate of AgCl proves chlorate. 

4. Acetate. Acidify the second portion of the filtrate from step 2 
with concentrated H 2 S0 4 , add a small volume of ethyl alcohol, and 
heat. 

Pleasant fruity odor of 6thyl acetate proves acetate. 



GROUP IV ANIONS 257 

COMBINED ANALYSIS OF THE ANIONS 

1. Test for Group I (the volatile acids) with dilute HC1. If 
present, proceed as in Group I scheme. 

2. Test a sample of the unknown, from which Group I has been 
eliminated with HC1, for Group II with BaCl 2 -CaCl 2 reagent. If a 
precipitate is obtained, proceed as in the group procedure. 

3. Acidify a fresh sample of the unknown with dilute HNO 3 , and 
add AgNOa solution. If Group III is present, proceed as previously 
directed. 

4. Treat another fresh portion with solid Ag 2 SO 4 or Ag2COs to 
remove most of the interferences (i.e., Groups I, II, and III). Filter, 
and test the nitrate for Group IV. 



CHAPTER XVI 
COMPLETE ANALYSIS 

The analysis of a complete unknown not only requires the applica- 
tion of all the ability and knowledge gained in the previous sections but 
also introduces some additional difficulties. The sample to be 
analyzed, usually a solid, can be a mixture of salts, a mineral, an ore, an 
alloy, or some definite commercial product, such as a paint pigment or a 
cleaning powder. After applying cation as well as anion analysis, the 
analyst should present a report that gives a qualitative but accurate 
account of the entire inorganic composition of the material. The 
analyst should also give an estimate of the amount of each substance 
present; i.c. r he should be able to state whether the constituents are 
present in large, medium, or small amounts. For instance, an analysis 
of Duralumin should read: 

Large amount of aluminum. 

Medium amount of copper. 

Small amounts of magnesium and iron. 

Let us now consider briefly the additional difficulties that we may 
encounter in the course of a complete analysis. 

Solution. The first problem is to dissolve the substance. This 
often demands a great deal of patience arid ingenuity. Certain steels 
are only so slowly soluble in acids that the student is likely to become 
discouraged. He should remember that success in the laboratory 
often requires the patience of a Thomas Edison. Certain types of 
substances can be dissolved only by special methods. These methods 
are discussed in the latter part of this section. 

Incompatibilities. Another complication is caused by the inter- 
action of some components of the mixture whenever the substance is 
dissolved or acidified. Let us assume that we wish to analyze an 
" effervescent powder" like Citro-carbonate. Upon dissolving this 
substance in water, a gas would be evolved. By means of the pre- 
liminary tests, we should find that the gas was carbon dioxide. We 
could therefore make the following deductions: (1) The original 
material must have contained a carbonate or bicarbonate. (2) Some 
substance must have been present originally that, upon the addition 
of water, gave an acidic solution. We should now analyze the solution 

258 



COMPLETE ANALYSIS 259 

and find that it contained sodium citrate. Hence we may say that the 
original substance was a mixture of citric acid and sodium carbonate or 
bicarbonate. Here we have chosen a very simple example, but in 
certain instances it is very difficult to decide what the original com- 
ponents were. 

Interferences. The presence of certain substances seriously inter- 
feres with the usual course of an analysis. For instance, tartrates and 
many other organic substances form very stable complexes with iron, 
aluminum, and chromium. These complexes furnish only minute 
amounts of the free cations and hence these ions cannot be precipitated 
by the usual methods. Consequently, before making a Group III 
cation analysis, the organic substance should be destroyed. Phos- 
phates and oxalatos also interfere with the analysis of cation Groups 
III and IV by forming precipitates of alkaline earth phosphates or 
oxalates when the solution is made alkaline. These interferences 
must therefore be removed before analyzing for Group III. 

Suggested Order of Analysis. The order in which the various 
steps of an analysis are made depends to a certain extent upon the 
substance. However, the scheme outlined has proved useful generally. 

1. Preliminary Tests. The unknown should be subjected to the 
series of tests that are outlined in the section on blowpipe analysis. 
These tests very often will rapidly show the analyst how he may 
materially shorten the course of an analysis. They also invariably 
give valuable indications as to the principal constituents in the 
unknown. At this point, the analyst should make a thorough tost for 
the presence of cyanides and carbonates, since it is obviously impossible 
to detect them in the prepared solution. 

2. Prepared Solution. The unknown should be boiled with sodium 
carbonate solution. (See details in later section.) This operation 
yields a solution of the sodium salts of the anions and precipitates the 
interfering cations. 

3. Analysis of the Anions. The prepared solution should be 
analyzed for the anions. 

4. Preparation of a Solution Suitable for Cation Analysis. See 
details in later section. 

5. Cation Analysis. If the preliminary tests and the anion analyses 
indicate the presence of interferences, they should be removed before 
the analysis of Group III is undertaken. 

Prepared Solution. The unknown is boiled with concentrated 
Na 2 CO 3 solution. The metals arc precipitated as the insoluble 
carbonates and hydroxides, and the filtrate contains the anions plus 
excess carbonate. 



260 QUALITATIVE ANALYSIS 

Preparation of a Solution Suitable for Cation Analysis. All inor- 
ganic substances can be classified in either of two groups. 

A. Water and Acid-soluble Substances. The finely ground material 
is treated with separate portions of cold and hot water, dilute and con- 
centrated hydrochloric acid, dilute and concentrated nitric acid. The 
order is determined by the following considerations. 

ADVANTAGES OF CONCENTRATED HC1 AS A SOLVENT. 

1. HC1 is a nonoxidizing acid. 

2. Metallic oxides dissolve more readily in HC1 than in HNOs. 

3. Oxidizing substances, when heated with HC1, evolve chlorine 
and hence indicate their presence. 

DISADVANTAGES OF HC1 AS A SOLVENT. 

1. Arsenous compounds are converted to volatile AsCls and hence 
may be lost. 

2. Silver and mercurous mercury are precipitated as insoluble 
chlorides. 

ADVANTAGES OF HN0 3 AS A SOLVENT. 

1. Excellent solvent for metals, alloys, and metallic sulfides. 

DISADVANTAGES OF HN0 3 AS A SOLVENT. 

1. Yields insoluble precipitates of metastannic acid and hydrated 
antimony oxides with compounds containing tin and antimony. 

2. Hot concentrated HN0 3 oxidizes sulfides and sulfites to sulfates 
and thus causes partial precipitation of the alkaline earth sulfates. 

Solution of Alloys. Small portions of the alloy should be treated 
with various solvents to determine which is most effective. The best 
solvents for the usual types of alloys are listed below. 

Copper alloys (brass, bronze) .... HNO 8 

Steels HC1 

Aluminum alloys HC1 

Silver alloys HNO 3 

Gold, platinum, and noble metals Aqua regia 

White metals Boiling HC1, to which HNO 3 is 

added, several drops at a time. 

This prevents the precipitation 

of metastannic acid 
Tin alloys Do not use HNO 3 except as given 

above 
High silicon ferrous alloys Hot 1 : 9 HC1 and subsequent 

concentration 

The only anions likely to be found in solutions of alloys are phos- 
phate, silicate, and sulfate, which are present originally as phosphides, 
silicides, and sulfides. Nitric acid is generally the preferred solvent, 
since it oxidizes these elements to the higher valence states. 



COMPLETE ANALYSIS 261 

B. Substances Insoluble in Water and Acids. If after all attempts 
to dissolve the substance in acid have been made, an insoluble residue 
is left, the number of possibilities is quite limited. The application of 
the methods of blowpipe analysis will usually furnish us with a tenta- 
tive identification. For conclusive proof, the special methods of 
solution, outlined below, should be applied and followed by identifica- 
tion tests. 

Alkaline Earth Sulfates. Fuse with a large excess of an anhydrous 
Na 2 C03-K 2 C03 mixture (1 : 1). After the mass becomes liquid, heating 
should be continued for 15 min. Allow to cool, and extract with 
dilute sodium carbonate solution until the sulfate ion is completely 
removed. The remaining alkaline earth carbonates arc dissolved in 
HC1. 

Lead Sulfate. Boil extensively with a concentrated sodium car- 
bonate solution, and wash the basic lead carbonate precipitate until the 
sulfate ion is completely removed. The residue is then dissolved in 
HN0 3 . 

Metastannic Acid and Tin Dioxide. Fuse with a large excess of a 
sulfur and anhydrous sodium carbonate mixture ^1:1). (Keep the 
crucible covered.) The residue is extracted wiMi warm water. The 
solution contains tin and other members of Groups IIB, IV, and V. 
The residue is dissolved in HNOs and tested for small amounts of 
members of Groups IIA and III. 

Silicic Acid and Silicates. Fuse with mixture of potassium and 
sodium carbonate. 

Insoluble Fluorides. Heating with concentrated sulfuric acid 
yields the corresponding alkaline earth sulfates. 

Silver Halides. Boiling with dilute sulfuric acid and granular 
zinc yields a black precipitate of metallic silver and a solution contain- 
ing the corresponding halide ion. 

Insoluble Chromium Compounds and Chromites. Fuse with 
Na 2 C0 3 and NaN0 3 . A yellow melt that dissolves in water yielding a 
yellow solution indicates the presence of chromate ion. 

Insoluble Complex Cyanides. Boil with strong NaOH solution, 
and dilute with water. The anions can be detected in the aqueous 
solution. 

INTERFERENCES IN CATION ANALYSIS 

Phosphate, oxalate, tartrate, and organic matter interfere with the 
analysis of cation Groups III and IV, either by forming precipitates 
when the solution is rendered alkaline (phosphate and oxalate) or by 
preventing the precipitation of certain ions by forming complex ions 



262 QUALITATIVE ANALYSIS 

(tartrate, citrate, etc.). These interferences must be tested for and 
removed, therefore, before analyzing for Group III. This necessitates 
modifying the scheme somewhat. 

Removal of Tartrate, Oxalate, and Organic Matter .Evaporate the 
filtrate from Group II just to dryncss, add concentrated H^Oi, and 
heat until the mass has charred completely. Cool, add a small volume 
of concentrated HN0 3 , and heat until fumes of S0 3 are evolved. Cool 
and repeat the treatment with HNOa until the solution becomes color- 
less or light straw color. (Usually three to five treatments are suffi- 
cient.) Dilute the solution, filter, and analyze the filtrate for Groups 
III and IV by the usual methods. The residue contains the sulfatcs of 
Ca, Ba, Sr, and Cr. Boil the residue for a short time with a solution of 
Na2COs. Filter and w^ash the precipitate until the washings are free 
from sulfate ion. Discard the filtrate and washings. Treat the pre- 
cipitate with dilute HC1, boil, and filter. Test the filtrate for Ca, Ba, 
Sr, and Cr. The sodium carbonate treatment transposes the insoluble 
sulfates to the insoluble carbonates. This transposition is not com- 
plete, particularly in the case of BaStKi, but yields sufficient ions so that 
tests may be performed. 

Removal of Phosphate. Boil the filtrate from Group II until it is 
free of H 2 S. Add a drop of HN0 3 , and boil to oxidize all the iron. 
Test a small sample of the solution for iron with K 4 Fe(CN) 6 . To the 
remainder of the solution, add ammonium hydroxide cautiously until a 
slight precipitate, which persists, is obtained, and then just redissolve 
in dilute HC1. Treat the solution with a quantity of solid NH^HsOa 
and some HC2H 3 2 . If the solution is not red at this point, add Feds 
solution until it assumes a deep red color. (Do not use an excess.) 

NOTE: The filtrate at this point should give a red precipitate with 
dilute NH 4 OH. A light-colored precipitate indicates that insufficient 
FeCls has been added. 

Dilute the solution with ten times its volume of hot water, and boil for 
several minutes. Filter while hot. The precipitate contains the 
phosphates arid basic acetates of Fe, Al, and Cr and possibly some Ni, 
Mn, and Zn. The colorless filtrate is concentrated to a small volume, 
and if a slight precipitate forms, it is filtered off and discarded. The 
filtrate is then subjected to analysis for the remainder of the cations. 
The precipitate is treated with solid Na 2 2 and boiled. The filtrate 
contains AlOr and CrC^", which are tested for in the usual manner. 



APPENDIX 
UNKNOWNS 

The analysis of unknowns is'thc central theme in a course in qualitative analysis. 
Hence it behooves us to give some attention to the mechanics involved in preparing 
and dispensing them. 

At the Polytechnic Institute, the following system has been used for a number of 
years and has been found to be very convenient and economical. Of course, many 
other systems can be and are used. To a large extent, the system that is chosen 
will depend upon the conditions existing in a particular institution. However, we 
are describing this system in the hope that it may bo of some value. 

Master Solutions of the Cations. A master solution is prepared of the nitrate 
of each of the cations analyzed for in the scheme (except those of Group Il/f, 
for which solutions of the chlorides are also prepared, and chromium, whose sulfato 
often may be used). This master solution 1ms a concentration equal to ten times 
the concentration of the test solutions (as given in the section on solutions). Usu- 
ally, 1 or 2 liters of each of these solutions is prepared. (From these master solu- 
tions, the test solutions may be readily prepared by dilution.) These solutions aro 
stabilized by adding the corresponding acids. They are kept in glass-stoppered 
bottles from year to year. 

Stock Solutions of Unknowns. From these master solutions, stock solutions of 
the unknowns are prepared. For instance, if we wished to prepare a stock solution 
containing Ag, On, As, Ni, Co, (\i, and Na, we should add 100 ml. of the master 
solution of each cation to a liter bottle and fill the bottle with distilled w.'iter, acid, 
or both. These stock solutions are serially numbered from one up (at least two 
labels to each bottle), and the key to the cations in each bottle is kept in the posses- 
sion of the instructor. There is no way of discovering from the number on a bottle 
what any unknown contains, unless one possesses the unknown book. These stock 
solutions are usually stable and are kept from year to year. Some are prepared 
annually, some biannually, some triannually, etc., as the solution becomes 
exhausted. 

Dispensing Unknowns. During the summer or shortly before the fall semester 
begins, a number of kits of unknowns is prepared, sufficient for the following year. 
Each student, upon registering in the laboratory, receives a kit containing the 
mhiimum number of unknowns which he must complete, and his name is entered 
on the appropriate page in the book. (Extra unknowns are dispensed individually 
as needed.) Each kit bears on the box a serial number that refers to a page in the 
record book of the instructor. The unknowns themselves arc numbered 1, 2, 3, 
etc. These numbers indicate the order in which the unknowns are to be per- 
formed. The student is told that unknown 1 contains Group I; number 2, Group 
HA; number 3, Group II/?; number 4, Groups I and II, etc. The student at no 
time knows the serial number of a particular unknown. Hence shopping around by 
the student in order to find someone who has the same number is eliminated. If the 
system described below is used in filling the kits, the probability that two kits will 
contain the same unknowns is extremely slight. 

263 



264 QUALITATIVE ANALYSIS 

The Kits. The kits themselves are small cardboard boxes that can be made to 
order by any manufacturer of cardboard boxes. They are not returnable and are 
usually discarded by the student at the end of the year. The liquid unknowns are 
dispensed in 8-dram (short style, patent lip) vials and stoppered with corks. Each 
vial holds about 30 ml., but the vials are not completely filled. Hence each student 
gets about 20 to 25 ml. of unknown, which is sufficient for from five to eight analy- 
ses. Thus the need for refills is largely eliminated. Our practice at the Poly- 
technic Institute has been to stamp the number of the unknowns with a rubber 
stamp on the cork of each bottle. We find this method much faster than pasting 
labels on the bottles. The number of mix-ups due to the interchanging of corks is, 
we find, negligible. 

Filling the Kits. The method described below for filling the kits is, we believe, 
almost foolproof; i.e., in the last five years that we have been using it, there has 
been but one case where a student received an unknown whose serial number was 
not correctly inscribed in the record book. The method has thus practically 
eliminated the common complaint, "I got the wrong unknown." 

The method used can be outlined thus: 

1. A record book is obtained with numbered pages. 

2. The kit boxes are numbered to correspond to the pages of the book; i.e., 
if the kit were numbered 42-34, the unknowns contained therein would have their 
serial numbers inscribed on page 34 of the 1942 record book. 

3. The kits .are then arranged in order of their numbers. 

4. All the liter bottles containing the solutions of the Group I unknowns are 
then placed on a laboratory bench. 

5. The required number of vials (equal to the number of kits) is counted out, 
and they are distributed among the liter bottles; i.e., if there are 5 Group 
I unknowns and 50 kits, 50 vials are counted out and 10 placed in front of each 
bottle. 

6. The vials in front of each bottle are filled from that bottle and replaced in 
front of that bottle. 

7. The vials are stoppered. 

8. Every cork is stamped with the number 1. 

9. The vials in front of the first bottle are then distributed among the kits. 
As each vial is placed in a kit box by an assistant, the serial number of the unknown 
is recorded on the appropriate page of the record book the number of the page is 
the same as the number of the kit by another person. (This operation is accom- 
plished most expeditiously by two persons.) 

10. When the vials in front of bottle 1 are distributed, the bottle containing 
that unknown is set aside, and the vials in front of bottle 2 are distributed and so 
on, until all the vials containing the first unknown have been distributed. At this 
point, every kit should have one vial, and there should be no extra vials. 

11. After this check, the same procedure is repeated for the second unknown 
and so on, until all the kits have been filled. 

Solid Unknowns. The anion and blowpipe unknowns are dispensed in the form 
of solid salts in 2-dram vials (about three-fourths full). The same system is fol- 
lowed in distributing these unknowns. 

Preparation of Solid Unknowns. Prepared solid unknowns are kept in screw- 
cap bottles, each of which has a serial number. About J^ to 1 Ib. of each unknown 
is prepared at one time. (This will vary with the number of unknowns, number of 
unknowns in a given series, etc.) The various salts (ground, if necessary, in a corn 



APPENDIX 265 

mill) are sieved through a 20-mesh sieve. This must be done even for very fine 
powders, since many fine powders form lumps. The required amounts of each 
powder are then violently shaken together in a large bottle until they are thoroughly 
mixed. They are then transferred to the stock bottle. 

Conclusions. This method, although it seems involved and appears to be quite 
time-consuming, is, in practice, quite simple and routine. As an example, an 
assistant working 4 hr. a day and a student working 7 hr. a day for 10 days filled, 
labeled, and recorded 1,440 vials of liquids and 600 vials of solids. In addition, 
they prepared master solutions and unknown stock solutions as needed. 

During the course of the year, very little extra time is needed. Additional 
unknowns performed for extra credit must be filled, and very occasionally, refills are 
needed. These refills are very easily made, since all the instructor needs to do is 
to consult the record book to see what the serial number of the unknown is and then 
to fill the student's vial from the appropriate stock bottle. 



MINIMUM LIST OF EQUIPMENT FOR EACH STUDENT FOR CATION 

ANALYSIS 

12 test tubes, 10 by 75 mm. Pyrex 

2 test tubes, 13 by 100 mm. Pyrex 

1 test tube, 20 by 150 mm. Pyrex 

1 test tube, side arm 5 in. 

1 test-tube rack 

1 test-tube holder (spring clothespin) 

1 crucible, 15 ml. 

1 crucible, 5 ml. 

2 beakers, 250 ml. 

1 flask, 500 ml. (for wash bottle) 

1 graduate, 10 ml. 

2 microscope slides, 3 by 1 in. 
6 stirring rods, 3 mm. by 8 in. 

12 medicine droppers (4-in. glass part) 

1 spot plate 

1 microburner 

1 ring stand and 2 rings 

1 burette clamp 

1 piece platinum wire, 2 in. 

1 sand pan 

1 water bath 

1 piece rubber tubing, Jf Q by %4, 12 in. long 

1 piece rubber tubing, J^ by ?-, 24 in. long 

25 sheets filter paper, Whatman No. 1, 11 cm. 

100 strips each litmus paper, red and blue 

Glass tubing, 7-mm. soft glass 

Rubber and cork stoppers 

Absorbent cotton 

Aitch-Tu-Ess cartridges 

Reagent bottles 

Centrifuge (1 for each 10 to 12 students) 

Hand spectroscope (1 for each 25 students) 



266 QUALITATIVE ANALYSIS 

ADDITIONAL EQUIPMENT FOB BLOWPIPE AND ANION ANALYSIS 

1 lead dish 

1 glass square, 3 by 3 in., soft glass 

1 blast lamp (for class) 

1 Bunsen burner 

1 hammer 

1 anvil 

1 blowpipe 

1 blowpipe tip 

1 charcoal borer (knife point may be used) 

Charcoal blocks 

Plaster of Paris tablets 



SOLUTIONS 

* Indicates that the substance is used in the scheme, or more than five times in 
the Laboratory part. 

t Indicates that the substance is used from three to five times in the Laboratory 
part. 

t Indicates that the substance is used only once or twice in the Laboratory part. 

Indicates that the substance is used in the scheme, or more than five times in 
only the Anion and Blowpipe sections. 

|| Indicates that the substance is used less than five times in only the Anion and 
Blowpipe sections. 

Acetic Acid. 

concentrated: glacial, approximately 17N* 

dilute; 1 part acid to 2 parts H 2 O, approximately QN* 

solution ; 30 per cent ; prepare as needed 
Alizarin S Blue. 

solution: 0.5 g./liter of 50% acetone. Add 1 ml. concentrated acetic acid* 
Aluminon. 

solution: 0.1 % aqueous solution* 
Aluminum Dust. 

solid 
Aluminum Nitrate. 

test solution: 70 g. Al(NO 3 ) 3 -9H 2 O/liter solution; add 8 ml. of coned. HNO 8 * 
Ammonium Acetate. 

solution 3N: 231 g. /liter solution J 

saturated aqueous solution* 

solid* 
Ammonium. Benzoate. 

solution: 7% in water* 
Ammonium Carbonate. 

solution QN: 250 g. powdered commercial ammonium carbonate dissolved in 1 1. 
of QN NH 4 OH (commercial ammonium carbonate is NH 4 HCO3-NH 4 CO2NH2)t 
Ammonium Chloride. 

saturated aqueous solution I 

solid* 
Ammonium Chromate. 

solution 0.5.ZV: 38 g./liter of solution* 
Ammonium Hydroxide. 

concentrated about 15N* 

dilute GN: (2 parts NH 4 OH:3 parts H 2 O)* 
Ammonium Molybdate. 

solution: dissolve 50 g. of 85 % molybdic acid in a mixture of 120 ml. of H 2 O and 
70 ml. of NH 4 OH. Filter, add 30 ml. of HNO 3 , and cool. Add the solution, 
with constant ( stirring, to a, mixture of 200 ml. HNOs and 480 ml. of H 2 O. 
Filter or decant solution before use. * 

9R7 



268 QUALITATIVE ANALYSIS 

Ammonium Nitrate. 

test solution: 44.4 g. /liter solution* 
Ammonium Oxalate. 

solution 0.5AT: 35 g. /liter solution* 
Ammonium Persulfate. 

solid! 
Ammonium Polysulfide. 

solution: saturate 200 ml. of ice-cold coned. NH 4 OH with H 2 S. Add 200 ml. of 
coned. NII 4 OH. Dilute to 1 1., and digest for several hours with 25 g. of 
flowers of sulfur. Filter.* 
Ammonium Sulfate. 

solution 0.5JV: 33 g./liter solution* 
Ammonium Thiocyanate. 

solution: saturated solution in acetone (NOTE: inflammable reagent)* 
Amyl Alcohol (iso). 

as commercially bought* 
Aniline Sulfate. 

solution: 3% aqueous solution || 
Antimonous Chloride. 

test solution: 19 g./liter of dilute HC1* 
Antimonous Oxide. 

solid || 
Arsenic Nitrate. 

test solution: dissolve 15.3 g. As 2 O5 in a liter of dilute HNO 3 * 
Arsenous Chloride. 

test solution: dissolve 13.2 g. of As 2 O 3 in a liter of dilute HC1* 
Arsenous Oxide. 

solid || 
Arsenous Sulfide. 

solid || 

Asbestos Fiber*. 
Barium Chloride. 

solution IN: 122 g. BaCl 2 -2H 2 0/liter solution! 

solid 
Barium Chloride -Calcium Chloride Reagent. 

solution: 240 g. BaCl 2 -2H 2 O and 219 g. OaCl 2 -6H 2 0/liter of solution 
Barium Hydroxide. 

saturated aqueous solution! 
Barium Nitrate. 

test solution: 19.0 g. Ba(NO 3 ) 2 -2H 2 O/liter of solution* 
Barium Peroxide. 

solid || 
Benzidine (Benzidine Acetate). 

solution: 0.5 g. benzidine in 100 ml. coned, acetic acid and dilute to 11,* 
a-Benzoin Oxime. 

solution: 5% alcoholic solution* 
Bettendorf s Reagent. 

saturated solution of stannous chloride in concentrated HC1* 
Bismuth Chloride. 

solid || 



APPENDIX 269 

Bismuth Flux. 

solid: equal parts of KI and sulfur ground together* 
Bismuth Nitrate. 

test solution: 23.2 g. of Bi(NO) 8 -5H 2 O/liter of dilute HNO 8 * 
Boric Acid. 

solid || 

Bromine Water, 
saturated aqueous solution (some liquid bromine should be left at the bottom of 

the bottle)* 
Brucine (Alkaloid). 

solid || 
C'aiotheline. 

saturated aqueous solution. Preparation of reagent : dissolve 20 g. of dry brucine 
in 100 ml. of 5 AT HNO 3 in the cold; then warm to 50 to 60 C. for 15 min. Crys- 
tals of cacotheline separate out. Cool in ice water, and allow crystallization 
to become complete. Filter with suction, and" wash with N HN0 3 , acetone 
and ether. Yield should be about 22 g. of yellow powdery crystals* 
Cadmium Carbonate. 

solid || 
Cadmium Chloride. 

solid || 

Cadmium Nitrate. 
test solution: 27.5 g. Cd(NO 3 ) 2 -4H 2 O/liter of solution. Ude 3 ml. of coned. 

HNCV 
Calcium Acetate. 

solid || 
Calcium Carbonate. 

solid || 

Calcium Chloride. 

solution 0.5AT: use 55 g. CaCl 2 -6H 2 O/liter of solution* 
solid || 
Calcium Fluoride. 

solid || 
Calcium Nitrate. 

test solution: use 58.9 g. Ca(NO 3 ) 2 -4H 2 O/liter of solution* 
Carbon Tetrachloride. 

liquid as bought 
Carnot's Reagent. 

solution: (freshly prepared) 1 drop of Q.5N bismuth nitrate solution is mixed with 
2 to 3 drops of 0.5AT sodium thiosulfatc solution and 10 to 15 ml. of absolute 
alcohol added. (Any turbidity is removed by the careful addition of 
water) t 
Chromic Nitrate. 

test solution: use 77 g. of Cr(NO 3 ) 3 -9H 2 0/liter of solution* 
Chromic Sulfate. 
test solution: use 68.9 g. Cr 2 (S0 4 )3-18H 2 0/liter of solution. Add 10 ml. coned. 

H 2 S0 4 * 
Cinchonine. 

solution: dissolve 10 g. cinchonine in a liter of hot water containing some HNO 3 , 
cool, and add 20 g. KI* 



270 QUALITATIVE ANALYSIS 

Cobaltous Nitrate. 

test solution: use 49.4 g. Co(N0 8 ) 2 -6H 2 0/liter of solution. Add 8 ml. of HN0 8 * 

solution 2N: 291 g. Co(NO 8 ) 2 -6H 2 O/liter of solution. Add HNO 8 
Copper. 

foil* 
Cupric Acetate. 

solid || 
Cupric Chloride. 

solid || 
Cupric Nitrate. 

test solution: 38 g. Cu(N0 3 ) 2 -3H 2 0/liter of solution* 
Cupric Sulfate. 

solution O.Stf: 63 g. CuSO 4 -5H 2 O/liter of solution || 
Dimethyl Glyoxime. 

1 % alcoholic solution* 
Diphenylamine Acetate. 

solution: 1 g. diphenylamine dissolved in 100 g. glacial acetic acid* 
Diphenyl Carbazide. 

saturated alcoholic solution* 
Ethyl Alcohol. 

95% liquid 
Ethyl Ether. 

purified (U.S.P.)* 
Ferric Chloride. 

solution 3#: 270 g./liter of solution 
Ferric Nitrate. 

test solution: 72.4 g. Fe(NO 3 ) 3 -9H 2 O/liter of solution; use 20 ml. HNO a * 
Ferrous Ammonium Sulfate. 

solid 

test solution: prepare as needed from clean crystals J 
Ferrous Sulfate. 

use ferrous ammonium sulfatej 
Fluorescein Paper. 

paper: dip filter paper in a saturated solution of fluorescein in alcohol- water 

(1:1); dry || 
Hydrochloric Acid. 

concentrated: about 12N* 

dilute: 6^ (1:1)* 

solution 0.2A r : prepare as needed by dilution* 
Hydrogen Peroxide. 

3% U.S.P. solution* 
Iodine Solution. 

solution 0.01 AT: 1.3 g. of iodine/liter of solution containing some KI|| 
Lanthanum Nitrate. 

solution 5 %: 5 g. dissolved in 95 ml. H 2 O|| 
Lead Acetate. 

solution: 95 g. Pb(C 2 H 8 O 2 ) 2 -3H 2 O/liter of solution || 

paper: moisten piece of filter paper with reagent t 
Lead Nitrate. 

test solution: 16 g. Pb(N0 8 ) 2 /liter of solution; use 2 ml. HN<V 



APPENDIX 271 

solid 1 1 

solution: 165 g. /liter of solution! 
Magnesia Mixture. 

solution: Dissolve 52 g. MgCl 2 -6H 2 Q and 134 g. NH 4 C1 in water, add 350 ml. of 

coned. NH 4 OH, dilute to 1 1. Use only freshly filtered or decanted liquid* 
Magnesium Metal. 

ribbon* 
Magnesium Nitrate. 

test solution: 106 g. Mg(N0 3 ) 2 -6H 2 0/liter of solution* 
Manganese Dioxide. 

solid || 
Manganous Chloride. 

solution: saturated solution of MnCl 2 -4H 2 O in coned. HC1 (freshly prepared) 

solid 
Manganous Nitrate. 

test solution: 53 g. Mn(N0 3 ) 2 -6H 2 O/liter of solution or 69 g. of 75 % solution /liter 

(75 % solution is sold commercially) * 
Manganous Sulfate. 

solid (MnS0 4 4H 2 0) || 
Mercuric Chloride. 

solution QAN: 54 g. /liter of solution* 

solid || 
Mercuric Iodide. 

solid || 
Mercuric Nitrate. 

test solution: 16 g. Hg(N0 3 ) 2 -KH 2 O/liter of solution; use 17 ml. HN0 8 * 

solution: 10 g. Hg(NO 3 ) 2 -HH 2 O/100 g. H 2 O; use 1 ml. HNO 3 || 
Mercuric Oxide Yellow. 

solid || 
Mercurous Nitrate. 

test solution: 14 g. Hg 2 (NO 3 ) 2 -2H 2 O/litcr of solution; use 50 ml. HNO 8 * 
Methyl Orange. 

solution: 0.1 % aqueous solution|| 
Naphthylamine Acetate. 

solution: boil 0.2 g. of solid a-naphthylamine with 20 ml. of H 2 O, pour off color- 
less solution, and add to it 150 ml. of 2AT acetic acid 
Nickelous Nitrate. 

test solution: 49.5 g. of Ni(NO 3 ) 2 -6H 2 O/liter of solution; use 10 ml. of HNO 8 * 
Nitric Acid. 

concentrated: approx. I5N* 

dilute: approx. 6N (2:3)* 
p-Nitrobenzene-azo-resorcinol. 

solution 0.001 %: 0.001 g. dissolved in 100 ml. of 2# NaOH* 
a-Nitroso-0-naphthol. 

solid* 

solution: make fresh as needed; saturated solution in 2 drops of 50 % acetic acid* 
Oxalic Acid. 

solid || 
Paraffin. 

solid|| 



272 QUALITATIVE ANALYSIS 

Perchloric Acid. 

60% solution % 
Phenolphthalein. 

1 % alcoholic solution || 
Picric Acid. 

saturated aqueous solution J 
Potassium Acetate. 

solid 
Potassium Bromide. 

solid || 
Potassium Chlorate. 

solid * 
Potassium Chloride. 

solid J 
Potassium Chromate. 

solution: 5 % aqueous solution* 

solid || 
Potassium Cyanide. 

solution: 3%* 

solid 1| 
Potassium Bichromate. 

solid || 
Potassium Ferricyanide. 

solution: 5% aqueous* 

solid || 
Potassium Ferrocyanide. 

solution: always freshly prepared from clean crystals* 

solid * 
Potassium Hydroxide. 

solid || 
Potassium Iodide. 

solid || 

solution IN: 166 g. /liter of solution* 

solution O.IN: prepare by dilution as needed J 
Potassium Nitrate. 

test solution: 26 g./liter of solution* 

solid || 
Potassium Nitrite. 

solid* 
Potassium Permanganate. 

solution Q.IN: 3.2 g./liter of solution* 
Potassium Thiocyanate. 

solution: 97 g./liter of solution* 
Pyridine. 

liquid t 
Rhodamine B. 

solution: 0.1 g. of tetraethylrhodamine /liter of solution* 
Rhodanine Solution. 

solution: 0.3 g. of p-dmiethylamino benzalrhodanine in 1 1. of acetone* 
Sand. 

pure quartz sand, "Ottawa sand"|| 



APPENDIX 273 

Silver Carbonate. 

solid || 
Silver Nitrate. 

test solution: 15.8 g./liter of solution* 

solution 0.1JV: use test solution* 

solution 50%: 50 g. AgNO 3 /50 ml. of H 2 0t 

solution 0.05%: prepare by diluting 50% solution* 

solid J 
Silver Sulfate. 

solid || 
Sodium Acetate. 

solid* 

saturated solution* 
Sodium Ammonium Hydrogen Phosphate (Microcosmic Salt). 

solid || 
Sodium Arsenate. 

test solution: 22.5 g. of Na 2 HAsO 4 -7II 2 O/liter of solution* 

solid 
Sodium Arsenite. 

solid 
Sodium Azide -Iodine Reagent. 

solution: in 100 nil. of 0.1 Af I 2 solution, dissolve 3 g. of sodium azidc|| 
Sodium Bicarbonate. 

solution O.OIW: 0.84 g./litcr of solution|| 

sol id 
Sodium Bismuthate. 

solid* 
Sodium Carbonate. 

solution 3N: 159 g./liter of solution* 

solution 0.01AT: 0.5 g./liter of solution|| 

solid anhydrous || 
Sodium Chloride. 

solution 5%: 50 g. in 950 ml. of H 2 O 

solid || 
Sodium Cobaltinitrite. 

solution: prepare fresh as needed* 

solid * 
Sodium Fluoride. 

solid || 
Sodium Hydroxide. 

solution QN: 240 g./liter of solution* 
Sodium Hypobromite. 

solution: prepare fresh as needed: to a solution of bromine water add NaOH until 

the solution is yellow; then add as much NaOH as was previously added* 
Sodium Nitrate. 

test solution: 37 g./liter of solution* 

solid || 
Sodium Nitrite. 

solid* 
Sodium Nitroprusside. 

solution: 1 g./lOO ml. of H a O 



274 QUALITATIVE ANALYSIS 

Sodium Oxalate. 

solid|| 
Sodium Peroxide. 

solid* 
prim. Sodium Phosphate NaH 2 PO 4 . 

solid || 
sec. Sodium Phosphate. 

solidH 

solution 10%: 100 g. of Na 2 HPO 4 '12H 2 O/900 g. of H 2 
tert. Sodium Phosphate. 

solid 1 1 
Sodium Potassium Tartrate. 

solid 1 1 
Sodium Stannite. 

solution: must be freshly prepared by the student; to a stannous chloride solu- 
tion add NaOH until the precipitate that forms redissolves (the solution should 
be alkaline) * 
Sodium sulfate. 

solid 1 1 
Sodium Sulfide. 

solid || 
Sodium Sulfite. 

solid 
Sodium Tetraborate (Borax). 

solid 
Sodium Thiosulfate. 

solution O.SAf: 62 g./liter of solution* 

solid 
Stannic Chloride. 

test solution: 29.5 g. SnCl 4 -5H 2 dissolved in 400 ml. of 6AT HC1 dilute to 1 1.* 
Stannous Chloride. 

test solution: 19.0 g. SnCl 2 -2H 2 dissolved in 1 1. of solution; use 100 ml. coned 
HC1* 

solution Q.5N: 56 g. dissolved in 1 1. of solution containing some HCl;add a piece 

of metallic tin* 
Starch-iodide Paper. 

paper can be purchased* 
Strontium Chloride. 

solution IN: 133 g. SrCl 2 -6H 2 O/liter of solution 

solid || 
Strontium Nitrate. 

test solution: 24.2 g. of Sr(N0 8 )2/liter of solution* 
Sulfanilic Acid. 

solution: 0.5 g. in 150 ml. of 2N acetic acid|| 
Sulfuric Acid. 

concentrated: about 36N* 

dilute :6AT (1:5)* 
Tartaric Acid. 

solid || 
Thiosinamine (Allyl Thiourea) 

solid* 



APPENDIX 275 

Tin. 

solid granulated*- 
Turmeric Paper. 

paper as bought || 
Zinc Metal. 

granular || 

stick (As free) * 
Zinc Chloride. 

solid || 
Zinc Nitrate. 

test solution: 45.5 g. Zn(NO 3 )2-6H 2 O/liter of solution* 
Zinc Nitroprusside. 

paste: prepare as needed; add concentrated zinc sulfate solution to a sodium 

nitroprusside solution; filter off the precipitate, and use as a paste 
Zinc Oxide. 

solid || 
Zinc Sulfate. 

solution: 25 g./lOO ml. H 2 
Zinc Uranyl Acetate. 

solution: dissolve 100 g. of uranyl acetate in 60 g. of 30 % acetic acid, warming, if 
necessary; dilute to 500 ml. (solution A); dissolve 300 g. of zinc acetate in 30 g. 
of 30 % acetic acid, and dilute to 500 ml. (solution B). Mix A and B\ add a 
trace of NaCl; allow to stand 24 hr.; filter* 



276 



QUALITATIVE ANALYSIS 



LOGARITHMS OF NUMBERS 1 



Natural I 
numbers.) 





1 


2 


3 


4 


5 


6 


7 


8 


9 


PROPORTIONAL PARTS. 


1 


r 

i 


3 


i 


p 


6 


tj 
i 


8 


9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 


8 


12 


17 


2 


25 


29 


33 


37 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


4 


8 


11 


IS 


19 


23 


26 


30 


34 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


3 


7 


10 


14 


17 


2 


24 


28 


31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


3 


6 


10 


13 


16 


19 


23 


26 


29 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


3 


6 


\ 


12 


15 


18 


21 


24 


27 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


3 


6 


g 


1 


14 


17 


20 


22 


25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


3 


5 


\ 


1 


13 


16 


18 


21 


24 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


2 


5 


' 


10 


12 


15 


17 


20 


22 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


2 


5 


' 


< 


12 


14 


16 


19 


21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


2 


4 


i 


i 


11 


13 


16 


18 


20 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2 


4 


( 




11 


13 


15 


17 


19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


2 


4 


6 




10 


12 


14 


16 


18 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


2 


4 


6 




10 


12 


14 


15 


17 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


2 


4 


6 


7 


< 


11 


13 


15 


17 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


2 


4 


5 


i 


1 


11 


12 


14 


16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


2 


3 


c 

1 





c 


10 


12 


14 


15 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2 


3 


5 


1 


8 


10 


11 


13 


15 


27 


[314 


^330 


t346 


1362 


4378 


t392 


44QC 


4425 


4440 


445( 


2 


3 


5 


( 


^ 


( 


I] 


1< 


14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


2 


3 


I 


( 


J 


9 


11 


12 


XT: 

14 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


1 


3 


^ 


6 


< 


9 


10 


12 


13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


1 


3 


4 


6 


i 


9 


10 


11 


13 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


1 


3 


t 


6 


i 


8 


10 


11 


12 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


1 


3 


4 


f 


7 


8 


9 


11 


12 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


1 


3 


4 


r 


6 


8 


9 


10 


12 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


1 


3 


4 


5 


6 


8 


9 


10 


11 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


1 


2 


4 


5 


6 


7 


c 


10 


11 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


1 


2 


4 


5 


6 


7 


8 


10 


11 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


1 


2 


g 


5 


6 


7 


8 


9 


10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


1 


2 


3 


f 


6 


7 


8 


9 


10 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


1 


2 


g 


4 


5 


7 


8 


9 


10 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


1 


2 


3 


4 


5 


6 


8 


9 


10 


41 


6128 


6138 


6149 


6160 


6170 


>180 


6191 


6201 


6212 


6222 


1 


2 


3 


4 


5 


6 


7 


8 


9 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


1 


2 


3 


4 


5 


6 


7 


8 


9 


43 


6335 


6345 


6355 


6365 


6375 


385 


6395 


6405 


6415 


425 


1 


2 


3 


4 


5 


6 


7 


8 


9 


44 


6435 


6444 


6454 


6464 


6474 


484 


6493 


6503 


6513 


522 


1 


2 


3 


4 


5 


6 


7 


8 


9 


45 


6532 


6542 


6551 


6561 


6571 


580 


6590 


6599 


6609 


618 


1 


2 


3 


4 


5 


6 


7 


8 


9 


46 


6628 


6637 


6646 


6656 


6665 


675 


6684 


6693 


6702 


712 


1 


2 


3 


4 


5 


6 


7 


7 


8 


47 


6721 


6730 


6739 


6749 


6758 


767 


6776 


6785 


6794 


803 


1 


2 


3 


4 


5 


5 


6 


7 


8 


48 


6812 


6821 


6830 


6839 


6848 


857 


6866 


6875 


6884 


893 


L 


2 


3 


4 


4 


5 


6 


7 


8 


49 


6902 


6911 


6920 


6928 


6937 


946 


6955 


6964 


6972 


981 


1 


2 


3 


4 


4 


5 


6 


7 


8 


50 


990 


6998 


7007 


7016 


7024 


033 


7042 


7050 


7059 


067 




2 


3 


3 


4 


5 


6 


7 


8 


51 


076 


7084 


7093 


7101 


7110 


118 


7126 


135 


143 


152 


L 


2 


3 


3 


4 


5 


6 


7 


8 


52 


160 


7168 


7177 


7185 


7193 


202 


210 


218 


226 


235 




2 


2 


3 


4 


5 


6 


7 


7 


53 


243 


7251 


7259 


7267 


7275 


284 


7292 


7300 


308 


316 




2 


2 


3 


4 


5 


6 


6 


7 


54 


7324 


7332 


7340 


7348 


7356 


364 


7372 


7380 


7388 


396 


L 


2 


2 


3 


4 


5 


6 


6 


7 



i From MWard, E. B., "Physical Chemistry for Colleges: 9 



LOGARITHMS OF NUMBERS 



277 



LOGARITHMS OF NUMBERS 



Natural 
numbers. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


PROPORTIONAL PARTS. 


1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


1 


2 


2 


3 


4 


5 


5 


6 


7 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


1 


2 


2 


3 


4 


5 


5 


6 


7 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


1 


2 


2 


3 


4 


5 


5 


6 


7 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


1 


1 


2 


3 


4 


4 


5 


6 


7 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


1 


1 


2 


3 


4 


4 


5 


6 


7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


1 


1 


2 


3 


4 


4 


5 


6 


6 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


1 


1 


2 


3 


4 


4 


5 


6 


6 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


1 


1 


2 


3 


3 


4 


5 


6 


6 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


1 


1 


2 


3 


3 


4 


5 


5 


6 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


1 


1 


2 


3 


3 


4 


5 


5 


6 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


1 


1 


2 


3 


3 


4 


5 


5 


6 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


1 


1 


2 


3 


3 


4 


5 


5 


6 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


1 


1 


2 


3 


3 


4 


5 


5 


6 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


7376 


8382 


1 


1 


2 


3 


3 


4 


4 


5 


6 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


1 


1 


2 


2 


3 


4 


4 


5 


6 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


1 


1 


2 


2 


3 


4 


4 


5 


6 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


1 


1 


2 


2 


3 


4 


4 


5 


5 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


1 


1 


2 


2 


^ 


4 


4 


5 


5 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


1 


1 


2 


2 


3 


4 


4 


5 


5 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


1 


1 


2 


2 


3 


4 


4 


5 


5 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


1 


1 


2 


2 


3 


3 


4 


5 


5 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


1 


1 


2 


2 


3 


3 


4 


5 


5 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


1 


1 


2 


2 


3 


3 


4 


4 


5 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


1 


1 


2 


2 


3 


3 


4 


4 


5 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9026 


1 


1 


2 


2 


3 


3 


4 


4 


5 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


1 


1 


2 


2 


3 


3 


4 


4 


5 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


1 


1 


2 


2 


3 


3 


4 


4 


5 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


1 


1 


2 


2 


3 


3 


4 


4 


5 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


1 


1 


2 


2 


3 


3 


4 


4 


5 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


1 


1 


2 


2 


3 


3 


4 


4 


5 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


1 


1 


2 


2 


3 


3 


4 


4 


5 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


1 


1 


2 


2 


3 


3 


4 


4 


5 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 





1 


1 


2 


2 


3 


3 


4 


4 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 





1 


1 


2 


2 


3 


3 


4 


4 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 





1 


1 


2 


2 


3 


3 


4 


4 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 





1 


1 


2 


2 


3 


3 


4 


4 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 





1 


1 


2 


2 


3 


3 


4 


4 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 





1 


1 


2 


2 


3 


3 


4 


4 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 





1 


1 


2 


2 


3 


3 


4 


4 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 





1 


1 


2 


2 


3 


3 


4 


4 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 





1 


1 


2 


2 


3 


3 


4 


4 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 





1 


1 


2 


2 


3 


3 


4 


4 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 





1 


1 


2 


2 


3 


3 


4 


4 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 





1 


1 


2 


2 


3 


3 


4 


4 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 





1 


1 


2 


2 


3 


3 


3 


4 



INDEX 



Acetate ion, reactions of, 255 

Acidity and basicity, definition of, 102, 

103 

in aqueous solution, 103 
in solution of salts, 103, 109Jf. 

degree of, 114 

modern theory of, 99#". 
Acids or alkalis, preliminary tests with, 
221 

table of, 221, 222 
Adhesion, 89 

Aluminum ion, reactions of, 188 
Ammonium ion, reactions of, 210 
Analytical tables, use of, 16 
Anions, combined analysis of, 257 

Group I, analysis of, 232 

Group II, analysis of, 242 

Group III, analysis of, 251 

Group IV, analysis of, 256 
Antimony-cinnabar test, 181 
Antimony ion, reactions of, 179 
Apparatus, 157jf. 
Arsenate ion, reactions of, 236 
Arsenic ion, reactions of, 177 
Arsenite ion, reactions of, 237 
Arsenous ion, reactions of, 177 
Atomic numbers, 9 

B 

Barium ion, reactions of, 205 
Bead tests, 218, 219 

tables of, 219, 220 
Bettendorf's test, 179 
Bismuth ion, reactions of, 174 
Blank tests, 155 
Blowpipe analysis, 213jf. 
Bond strength, 104# 

in amphoteric substances, 106-109 

in hydro compounds, 105 

in hydroxy compounds, 106 

in oxy compounds, 104 



Borate ion, reactions of, 241 
Boyle's law, 57, 58 
Bromide ion, reactions of, 249 
Buffer solutions, 128 



Cadmium ion, reactions of, 176 
Calcium ion, reactions of, 203 
Calomel electrode, 146 
Carbonate ion, reactions of, 223 
Cations, Group I, analysis of, 171 

table of, 172 
Group II, analysis of, 182 

tables of, 187, 188 
Group III, analysis of, 198 

tables of, 202, 203 
Group IV, analysis of, 206 

table of, 207 
Group V, analysis of, 210 

table of, 212 

Centrifuge, operation of, 158 
Charcoal block tests, 216, 217 

table of, 217 
Charles's law, 58 
Chelate compounds, 50 
Chlorate ion, reactions of, 254 
Chloride ion, reactions of, 248 
Chromate ion, reactions of, 239 
Chromium ion, reactions of, 191 
Cobaltous ion, reactions of, 194 
Common-ion effect, 126 
Complete analysis, 258^". 
Completeness of reaction due to, com- 
mon ion effect, 126-129 
formation of complex ion, 132-134 
of difficultly soluble precipitate, 

122-129 

of slightly ionized product, 129-132 
of volatile gas, 132 

Complex ions, application of, to ana- 
lytical procedures, 51 



279 



280 



QUALITATIVE ANALYSIS 



Complex ions and molecules, conduc- 
tivity of solutions of, 45 

nomenclature of, 53 

occurrence of, 38 

stability of, 45 

structure of, 39 

types of, 43 
Complexes, acido, 48 

ammono, 47 

aquo, 47 

hydro, 47 

inner, 50 

poly acido, 49 

polyhalido, 49 

polysulfido, 49 

Compounds, complex, nomenclature of, 
53 

three-element, nomenclature of, 35 

two-element, nomenclature of, 34 
Control tests, 155 
Coordination number, 41 
Copper, reactions of, 175 
Covalent bonds, 32 
Crystal structure and X rays, 62-64 
Crystallization, 94 
Crystals, lattice types of, 65 

space geometry of, 64, 65 

structure of, 61-65 
Cupric ion, reactions of, 175 
Cyanide ion, reactions of, 230 

D 

Dalton's law, 58 

Dichromate ion, reactions of, 239 

Dipoles, 32, 42 

induced, 78 

rigid, 77 
Droppers, construction of, 160 

E 

Effective atomic number, 46 
Elements, in cation Group I, 17 

Group II, 17 

Group IIA, 18 

Group IIB, 18 

Group III, 19 

Group IIIA, 19 

Group IIIB, 20 

Group IV, 20 

Group V, 21 



Elements, symbolic pictures of, 2&ff. 

in usual cation scheme, 15 
Electron distribution of the elements, 

10 

Electrovalent bonds, 32 
Energy, chemical, 75 
free, 75 
thermal, 75 
work, 74 
Equilibrium constants, in terms of 

concentration, 78-85 
correlation of, in terms of pressure, 

85 

of hydrogen iodide, 81-83 
of phosphorus pentachloride, 80 
in terms of pressure, 85-88 
of hydrogen iodide, 87 
of phosphorus pentachloride, 85 
Equipment, minimum list of, 265, 266 
Extraction, methods of, 161 

F 

Ferric ion, reactions of, 189 
Ferricyanidc ion, reactions of, 246 
Ferrocyanide ion, reactions of, 244 
Ferrous ion, reactions of, 189 
Filtration, methods of, 159 
Flame tests, 220 

table of, 221 
Fluoride ion, reactions of, 233 

G 

Gas generator, construction of, 224, 225 
Gas laws, 57ff. 

experimental verification of, 59 

combined, 59 
Gases, kinetic theory of, 55jf. 

reactions between, 78-89 
Glass blowing, 224, 225 
Group separations, table of, 212 
Gutzeit test, 179 

H 

Hepar test, 217 
Heteropolar bond, 31, 77 
Homopolar bond, 31, 76 
Hydrogen atom, 24 ff. 

Bohr picture, 24 

Born concept, 26 

Schroedinger and de Broglie concept, 
26 



INDEX 



281 



Hydrogen electrode, 145 

Hydrogen molecule, 30 

Hydrogen sulfide, generation of, 162 



Ideal gas, 59 
Incompatibilities, 258 
Insoluble substances, 261 
Instability constant, 134 
Interferences, 259 
Iodide ion, reactions of, 250 
lonization constants, table of, 113 

of water, table of, 115 
Iron, reactions of, 189 
Isomorphism, 66 

K 

Kinetic theory, 55jf. 
L 

Laue diagrams, 63, 64 
Lead ion, reactions of, 169, 173 
Limit of detection, definition of, 6 
Liquids, 67, 68 

reactions between, 89-91 

structure of, 90 

M 

Manganous ion, reactions of, 196 
Magnesium ion, reactions of, 207 
Mass-action law, 121 
Matrass tests, 214-216 

table of, 215, 216 
Mercuric ion, reactions of, 172 
Mercurous ion, reactions of, 170 
Molar potentials, 146 

table of, 147 

N 

Nickel ion, reactions of, 193 

Nitrate ion, reactions of, 253 

Nitrite ion, reactions of, 231 

Nomenclature of complex ions and 

molecules, 53 

of three-element compounds, 35 
of two-element compounds, 34 



Nonideal gases, 60 

Notebook, the organization of, 155, 156 



O 



Order of analysis, suggested, 259 
Oxalate ion, reactions of, 240 
Oxidation-reduction, electronic explana- 
tion of, 137 

equations, balancing of, 140-144 
experimental basis of, 136, 137 
experimental evidence of, 138, 139 
Oxidizing power, measurement of the, 
144/. 



Periodic System, diagonal relationship 

in, 14 

horizontal relationships in, 13 
vertical relationships in, 12 
Periodic Tablo, 9 
Peter Griess test, 232 
pll, 115jf. 

Phosphate ion, reactions of, 235 
Phosphorus pentachloride, equilibrium 

constant of, 80, 85 
Plaster of Paris tablet tests, 218 

table, 218 

Potassium ion, reactions of, 208 
Prepared solution, 259 
Precipitates, transfer of, 161 

washing of, 160 

Precipitation, completeness of, 158 
Pure substances, the energy approach 

to reactions of, 73 
the kinetic approach to reactions of, 

70 

the governing condition to, 71 
mechanism of solution of, 92, 93 
oxidation-reduction of, in solution, 



reaction of, 70ff. 
reactions in solution, 121jf. 
reactions, with solvents, 92jf. 

R 

Reactions, the extent of, 73 
Reagent bottles, 163 



282 



QUALITATIVE ANALYSIS 



Reference electrode, 145 
Removal of cation interferences, 261, 
262 



Salt bridge, the 139 
Semipolar bond, the 31, 77 
Sensitivity, definition of, 5 
Silicate ion, reactions of, 242 
Silver ion, reactions of, 167 
Silver mirror test, 239 
Sodium chloride, structure of, 65 
Sodium ion, reactions of, 209 
Solids, 61-67 

amorphous, 61 

reactions between, 88, 89 
Solubility, 94 

degree of, 98 
Solubility calculations, 95 

molal solutions, 95 

molar solutionsf 95 

mole fraction, 95 

mole per cent, 95 

normal solutions, 96 

percentage solutions, 95 

specific gravity, 97 
Solubility product, calculation of, 124 

definition of, 123 

principle of, 122 
application of, 125 

table of, 123 
Solutions, of alloys, 260 

aqueous, 99-101 

experimental characteristics of, 98 

heating of, 161 

ionic, 99 

liquid ammonia, 101 

nonionic, 98 

preparation of, 267-275 

selenium oxychloride, 101 

suitable for cation analysis, prepa- 
ration of, 260 
Specificity, definition of, 5 
Spectroscope, hand, 165 



Spot tests, technique of, 164 
Stannic ion, reactions of, 181 
Stannous ion, reactions of, 181 
Strontium ion, reactions of, 204 
Sulfate ion, reactions of, 234 
Sulfide ion, reactions of, 229 
Sulfite ion, reactions of, 226 



Tartrate ion, reactions of, 238 

Technique, 157jf. 

Thermodynamics, the first law of, 73, 74 

the second law of, 74 
Thiocyanate ion, reactions of, 247 
Thiosulfate ion, reactions of, 228 



U 



Unknown sample, solution of, 258 
Unknowns, preparation and dispensing 
of, 263^. 



Valence forces, primary, 32, 76, 77 

polar bond, 31, 77 

homopolar bond, 31, 76 

semipolar bond, 31, 77 
secondary, 77, 78 

dispersion forces, 78 

induced dipolcs, 78 

rigid dipoles, 77 
Valence number, 33 
van der Waals' equation, 60 

W 

Werner's theory, modern interpreta- 
tion of, 41 
original concept of, 39 



Zinc ion, reactions of, 197